Limit (BCB-Revised Edition 2020, Exercise 2, Page 379)

Evaluate the following.

1. \( \displaystyle \lim_{x \to 0} \frac{\sin ax}{x} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin ax}{x} \)
   
   or \( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} .a \)
   
   or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} \right ) .a \)
   
   or \( a\)
   
   This completes the solution

2. \( \displaystyle \lim_{x \to 0} \frac{\tan bx}{x} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\tan bx}{x} \)
   
   or \( \displaystyle \lim_{x \to 0} \frac{\tan bx}{bx} .b \)
   
   or \( \left ( \displaystyle \lim_{x \to 0} \frac{\tan bx}{x} \right ) .b \)
   
   or \( b\)
   
   This completes the solution

3. \( \displaystyle \lim_{x \to 0} \frac{\sin mx}{\sin nx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin mx}{\sin nx} \)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\sin mx}{mx}}{\frac{\sin nx}{nx}} \right ) \frac{mx}{nx} \)
   
   or \( \left ( \frac{ \displaystyle \lim_{x \to 0} \frac{\sin mx}{mx}}{ \displaystyle \lim_{x \to 0} \frac{\sin nx}{nx}} \right ) \frac{m}{n} \)
   
   or \( \left ( \displaystyle \frac{ 1}{ 1} \right ) \frac{m}{n} \)
   
   or \( \frac{m}{n} \)
   
   This completes the solution

4. \( \displaystyle \lim_{x \to 0} \frac{\tan ax}{\tan bx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\tan ax}{\tan bx}\)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\tan ax}{ax}}{\frac{\tan bx}{bx}} \right ) \frac{ax}{bx} \)
   
   or \( \left ( \frac{ \displaystyle \lim_{x \to 0} \frac{\tan ax}{ax}}{ \displaystyle \lim_{x \to 0} \frac{\tan bx}{bx}} \right ) \frac{a}{b} \)
   
   or \( \left ( \displaystyle \frac{ 1}{ 1} \right ) \frac{a}{b} \)
   
   or \( \frac{a}{b} \)
   
   This completes the solution

5. \( \displaystyle \lim_{x \to 0} \frac{\sin px}{\tan qx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin px}{\tan qx}\)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\sin px}{px}}{\frac{\tan qx}{qx}} \right ) \frac{px}{qx} \)
   
   or \( \left ( \frac{ \displaystyle \lim_{x \to 0} \frac{\sin px}{px}}{ \displaystyle \lim_{x \to 0} \frac{\tan qx}{qx}} \right ) \frac{p}{q} \)
   
   or \( \left ( \displaystyle \frac{ 1}{ 1} \right ) \frac{p}{q} \)
   
   or \( \frac{p}{q} \)
   
   This completes the solution

6. \( \displaystyle \lim_{x \to a} \frac{\sin (x-a)}{x^2-a^2} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to a} \frac{\sin (x-a)}{x^2-a^2}\)
   
   or \( \displaystyle \lim_{x \to a} \frac{\sin (x-a)}{x-a} \frac{1}{x+a}\)
   
   or \( \left ( \displaystyle \lim_{(x-a) \to 0} \frac{\sin (x-a)}{(x-a)} \right ) \left ( \displaystyle \lim_{x \to a} \frac{1}{x+a} \right )\)
   
   or \( \left ( 1 \right ) \left ( \frac{1}{a+a} \right ) \)
   
   or \( \frac{1}{2a} \)
   
   This completes the solution

7. \( \displaystyle \lim_{x \to p} \frac{x^2-p^2}{\tan (x-p)} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to p} \frac{x^2-p^2}{\tan (x-p)}\)
   
   or \( \displaystyle \lim_{x \to p} (x+p) \frac{(x-p)}{\tan (x-p)}\)
   
   or \( \displaystyle \lim_{x \to p} (x+p) \left ( \displaystyle \lim_{(x-p) \to 0} \frac{(x-p)}{\tan (x-p)} \right ) \)
   
   or \( (p+p) \left ( 1\right )\)
   
   or \(2p \)
   
   This completes the solution

8. \( \displaystyle \lim_{x \to 0} \frac{\sin ax. \cos bx}{\sin cx} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{\sin ax. \cos bx}{\sin cx} \)
   
   or \( \displaystyle \lim_{x \to 0} \left ( \frac{ \frac{\sin ax}{ax} (\cos bx) }{\frac{\sin cx}{cx} } \right ) \frac{(ax)}{(cx)}\)
   
   or \( \frac{ \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} \right ) \left ( \displaystyle \lim_{x \to 0} \cos bx \right ) }{ \left ( \displaystyle \lim_{x \to 0} \frac{\sin cx}{cx} \right ) } \frac{a}{c} \)
   
   or \( \frac{ \left ( 1 \right ) \left ( 1 \right ) }{ \left ( 1 \right ) } \frac{a}{c} \)
   
   or \( \frac{a}{c}\)
   
   This completes the solution

   ---

9. \( \displaystyle \lim_{x \to 0} \frac{1-\cos x}{x^2} \)
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 0} \frac{1-\cos x}{x^2} \)
   
   or \( \displaystyle \lim_{x \to 0} \frac{1-\cos x}{x^2} . \frac{1+\cos x}{1+\cos x}\)
   
   or \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 x}{x^2} . \frac{1}{1+\cos x}\)
   
   or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 x}{x^2} . \frac{1}{1+\cos x}\)
   
   or \( \displaystyle \lim_{x \to 0} \left (\frac{\sin x}{x}\right )^2 . \frac{1}{1+\cos x}\)
   
   or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin x}{x}\right )^2 . \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos x} \right )\)
   
   or \( \left ( 1 \right )^2 . \left ( \frac{1}{1+1} \right )\)
   
   or \( \frac{1}{2}\)
   
   This completes the solution

10. \( \displaystyle \lim_{x \to 0} \frac{1-\cos 6x}{x^2} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{1-\cos 6x}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos 6x}{x^2} . \frac{1+\cos 6x}{1+\cos 6x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 6x}{x^2} . \frac{1}{1+\cos 6x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 6x}{x^2} . \frac{1}{1+\cos 6x}\)
    
    or \( \displaystyle \lim_{x \to 0} \left (\frac{\sin 6x}{6x}\right )^2 .6^2. \frac{1}{1+\cos 6x}\)
    
    or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin 6x}{6x}\right )^2 .36. \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos 6x} \right )\)
    
    or \( \left ( 1 \right )^2 .36. \left ( \frac{1}{1+1} \right )\)
    
    or \( 18\)
    
    This completes the solution

11. \( \displaystyle \lim_{x \to 0} \frac{1-\cos 9x}{x^2} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{1-\cos 9x}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos 9x}{x^2} . \frac{1+\cos 9x}{1+\cos 9x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 9x}{x^2} . \frac{1}{1+\cos 9x}\)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 9x}{x^2} . \frac{1}{1+\cos 9x}\)
    
    or \( \displaystyle \lim_{x \to 0} \left (\frac{\sin 9x}{9x}\right )^2 .9^2. \frac{1}{1+\cos 9x}\)
    
    or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin 9x}{9x}\right )^2 .81. \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos 9x} \right )\)
    
    or \( \left ( 1 \right )^2 .81. \left ( \frac{1}{1+1} \right )\)
    
    or \( \frac{81}{2}\)
    
    This completes the solution

12. \( \displaystyle \lim_{x \to 0} \frac{\cos ax-\cos bx}{x^2} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\cos ax-\cos bx}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\cos ax-\cos bx}{x^2} . \frac{\cos ax+\cos bx}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\cos^2 ax-\cos^2 bx}{x^2} . \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{(1-\sin^2 ax)-(1-\sin^2 bx)}{x^2} . \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\sin^2 bx-\sin^2 ax}{x^2} . \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \left ( \frac{\sin^2 bx}{x^2} -\frac{\sin^2 ax}{x^2} \right ). \frac{1}{\cos ax+\cos bx} \)
    
    or \( \displaystyle \lim_{x \to 0} \left ( \left ( \frac{\sin bx}{bx}\right )^2 .b^2- \left ( \frac{\sin ax}{ax} \right )^2.a^2 \right ). \frac{1}{\cos ax+\cos bx} \)
    
    or \( \left ( \left ( \displaystyle \lim_{x \to 0} \frac{\sin bx}{bx}\right )^2 .b^2- \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax} \right )^2.a^2 \right ). \left ( \displaystyle \lim_{x \to 0} \frac{1}{\cos ax+\cos bx} \right ) \)
    
    or \( \left ( 1)^2 .b^2- \left ( 1 \right )^2.a^2 \right ). \left ( \frac{1}{1+1} \right ) \)
    
    or \( \frac{b^2- a^2}{2} \)
    
    This completes the solution

13. \( \displaystyle \lim_{x \to 0} \frac{\sin ax-\sin bx}{x} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\sin ax-\sin bx}{x} \)
    
    or \( \displaystyle \lim_{x \to 0} \left ( \frac{\sin ax}{x}-\frac{\sin bx}{x} \right ) \)
    
    or \( \left ( \displaystyle \lim_{x \to 0} \frac{\sin ax}{ax}. a- \displaystyle \lim_{x \to 0} \frac{\sin bx}{bx}.b \right ) \)
    
    or \( \left ( 1.a- 1.b \right ) \)
    
    or \( a-b \)
    
    This completes the solution

14. \( \displaystyle \lim_{x \to 0} \frac{1-\cos px}{1-\cos qx} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{1-\cos px}{1-\cos qx} \)
    
    or \( \displaystyle \lim_{x \to 0} (1-\cos px)\frac{1}{1-\cos qx} \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( (1-\cos px). \frac{1+\cos px}{1+\cos px} \right ) \left ( \frac{1}{1-\cos qx} .\frac{1+\cos qx}{1+\cos qx} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{1-\cos^2 px}{1+\cos px} \right ) \left ( \frac{1+\cos qx}{1-\cos^2 qx} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{\sin^2 px}{1+\cos px} \right ) \left ( \frac{1+\cos qx}{\sin^2 qx} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{\sin^2 px}{\sin^2 qx} \right ) \left ( \frac{1+\cos qx}{1+\cos px} \right ) \)
    
    or \( \displaystyle \lim_{x \to 0}\left ( \frac{\frac{\sin px}{px} }{\frac{\sin qx}{q}} . \frac{px}{qx}\right )^2 \left ( \frac{1+\cos qx}{1+\cos px} \right ) \)
    
    or \( \left ( \frac{\displaystyle \lim_{x \to 0} \frac{\sin px}{px}}{\displaystyle \lim_{x \to 0} \frac{\sin qx}{q}} . \frac{p}{q}\right )^2 \left (\displaystyle \lim_{x \to 0} \frac{1+\cos qx}{1+\cos px} \right ) \)
    
    or \( \left (\frac{p}{q}\right )^2 \left (\frac{1+1}{1+1} \right ) \)
    
    or \( \frac{p^2}{q^2}\)
    
    This completes the solution

---

15. \( \displaystyle \lim_{x \to 0} \frac{\tan x-\sin x}{x^3} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\tan x-\sin x}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x (1-\cos x)}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \frac{(1-\cos x)}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \frac{(1-\cos^2 x)}{x^2} \frac{1}{1+\cos x} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \frac{(\sin^2 x}{x^2} \frac{1}{1+\cos x} \)
    
    or \(\left ( \displaystyle \lim_{x \to 0} \frac{\tan x}{x} \right ) \left ( \displaystyle \lim_{x \to 0} \frac{\sin x}{x} \right )^2 \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos x} \right ) \)
    
    or \(1.1.\frac{1}{1+1} \)
    
    or \(\frac{1}{2} \)
    
    This completes the solution

16. \( \displaystyle \lim_{x \to 0} \frac{\tan 2x-\sin 2x}{x^3} \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 0} \frac{\tan 2x-\sin 2x}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x (1-\cos 2x)}{x^3} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{x} \frac{(1-\cos 2x)}{x^2} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{x} \frac{(1-\cos^2 2x)}{x^2} \frac{1}{1+\cos 2x} \)
    
    or \( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{2x}.(2) \frac{\sin^2 2x}{(2x)^2} .(2^2) \frac{1}{1+\cos 2x} \)
    
    or \(\left ( \displaystyle \lim_{x \to 0} \frac{\tan 2x}{2x} .(2)\right ) \left ( \displaystyle \lim_{x \to 0} \frac{\sin 2x}{2x} .(2) \right )^2 \left ( \displaystyle \lim_{x \to 0} \frac{1}{1+\cos 2x} \right ) \)
    
    or \(1(2).1.(4)\frac{1}{1+1} \)
    
    or \(4 \)
    
    This completes the solution

17. \( \displaystyle \lim_{x \to \frac{\pi}{2}} (\sec x -\tan x) \)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \frac{\pi}{2}} (\sec x -\tan x) \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \sec x (1 -\sin x) \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sec x (1 -\sin^2 x)}{1+\sin x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sec x \cos^2 x}{1+\sin x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{1+\sin x} \)
    
    or \( \frac{\cos \frac{\pi}{2}}{1+\sin \frac{\pi}{2}} \)
    
    or \( \frac{0}{1+1} \)
    
    or \( 0 \)
    
    This completes the solution

18. \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sec ^2 x-2}{\tan x-1}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sec ^2 x-2}{\tan x-1} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{1+\tan^2 x-2}{\tan x-1} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\tan^2 x-1}{\tan x-1} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \tan x+1\)
    
    or \( \tan \frac{\pi}{4}+1\)
    
    or \( 1+1=2\)
    
    This completes the solution

19. \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- \csc ^2 x}{1 -\cot x}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- \csc ^2 x}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- (1+\cot ^2 x)}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{2- 1-\cot ^2 x}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{1-\cot^2 x}{1 -\cot x} \)
    
    or \( \displaystyle \lim_{x \to \frac{\pi}{4}} 1+\cot x\)
    
    or \( 1+\cot \frac{\pi}{4}\)
    
    or \( 1+1=2\)
    
    This completes the solution

20. \( \displaystyle \lim_{x \to y} \frac{\tan x -\tan y}{x-y}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to y} \frac{\tan x -\tan y}{x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{\tan (x-y). (1+\tan x \tan y)} {x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{\tan (x-y)}{x-y}. (1+\tan x \tan y) \)
    
    or \( \displaystyle \lim_{(x-y) \to 0} \frac{\tan (x-y)}{x-y}. \displaystyle \lim_{x \to y} (1+\tan x \tan y) \)
    
    or \( 1. (1+\tan x \tan x) \)
    
    or \( 1. (1+\tan^2 x ) \)
    
    or \( \sec^2 x \)
    
    This completes the solution

21. \( \displaystyle \lim_{x \to y} \frac{\sin x -\sin y}{x-y}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to y} \frac{\sin x -\sin y}{x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{2 \cos (\frac{x+y}{2}) .\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} 2 \cos (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} 2 \cos (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{(\frac{x-y}{2})} . \frac{1}{2}\)
    
    or \( \left ( \displaystyle \lim_{x \to y} \cos \frac{x+y}{2} \right ) \left ( \displaystyle \lim_{x \to y} \frac{\sin \frac{x-y}{2}}{\frac{x-y}{2}} \right ) \)
    
    or \( \cos \frac{y+y}{2} ( 1 ) \)
    
    or \( \cos y\)
    
    This completes the solution

22. \( \displaystyle \lim_{x \to y} \frac{\cos x -\cos y}{x-y}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to y} \frac{\cos x -\cos y}{x-y} \)
    
    or \( \displaystyle \lim_{x \to y} \frac{-2 \sin (\frac{x+y}{2}) .\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} -2 \sin (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{x-y}\)
    
    or \( \displaystyle \lim_{x \to y} -2 \sin (\frac{x+y}{2}) \frac{\sin (\frac{x-y}{2})}{(\frac{x-y}{2})} . \frac{1}{2}\)
    
    or \( \left ( \displaystyle \lim_{x \to y} -\sin \frac{x+y}{2} \right ) \left ( \displaystyle \lim_{x \to y} \frac{\sin \frac{x-y}{2}}{\frac{x-y}{2}} \right ) \)
    
    or \( -\sin \frac{y+y}{2} ( 1 ) \)
    
    or \( -\sin y\)
    
    This completes the solution

23. \( \displaystyle \lim_{x \to \theta} \frac{x \cot \theta -\theta \cot x}{x -\theta}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \theta} \frac{x \cot \theta -\theta \cot x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{x \cot \theta - \theta \cot \theta + \theta \cot \theta - \theta \cot x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cot \theta + \theta (\cot \theta - \cot x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cot \theta} {x -\theta} + \displaystyle \lim_{x \to \theta} \frac{\theta (\cot \theta - \cot x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \cot \theta - \displaystyle \lim_{x \to \theta} \frac{\theta (\cot x -\cot \theta )}{x -\theta} \)
    
    or \( \cot \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ - \frac{\sin(x-\theta)}{\sin x \sin \theta} }{x -\theta} \)
    
    or \( \cot \theta + \theta \displaystyle \lim_{x \to \theta} \frac{\sin(x-\theta)}{x -\theta} \frac{1}{\sin x \sin \theta} \)
    
    or \( \cot \theta + \theta \left ( \displaystyle \lim_{x-\theta \to 0} \frac{\sin(x-\theta)}{x -\theta} \right ) \left ( \displaystyle \lim_{x \to \theta} \frac{1}{\sin x \sin \theta} \right ) \)
    
    or \( \cot \theta + \theta .1. \frac{1}{\sin^2 \theta} \)
    
    or \( \cot \theta + \theta \csc^2 \theta \)
    
    This completes the solution

24. \( \displaystyle \lim_{x \to \theta} \frac{x \cos \theta -\theta \cos x}{x -\theta}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \theta} \frac{x \cos \theta -\theta \cos x}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{x \cos \theta -\theta \cos \theta+\theta \cos \theta -\theta \cos x}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cos \theta+ \theta ( \cos \theta - \cos x)}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \cos \theta }{x -\theta} + \frac{ \theta (\cos \theta - \cos x)}{x -\theta}\)
    
    or \( \displaystyle \lim_{x \to \theta} \cos \theta + \theta \displaystyle \lim_{x \to \theta} \frac{ ( 2 \sin (\frac{x+\theta}{2}). \sin (\frac{x-\theta}{2}))}{x -\theta}\)
    
    or \( \cos \theta + \theta \left ( \displaystyle \lim_{x \to \theta} \sin (\frac{x+\theta}{2}) \right ) \left (\displaystyle \lim_{x-\theta \to 0} \frac{ \sin (\frac{x-\theta}{2})}{ (\frac{x-\theta}{2})} \right ) \)
    
    or \( \cos \theta + \theta \left ( \sin \theta \right ) \left (1 \right ) \)
    
    or \( \cos \theta + \theta \sin \theta \)
    
    This completes the solution

25. \( \displaystyle \lim_{x \to 1} \frac{1+ \cos \pi x}{\tan ^2 \pi x}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to 1} \frac{1+ \cos \pi x}{\tan ^2 \pi x}\)
    
    or \( \displaystyle \lim_{x \to 1} \frac{1- \cos (\pi-\pi x)}{\tan ^2 (\pi-\pi x)} \)
    
    or \( \displaystyle \lim_{x \to 1} \frac{1- \cos^2 (\pi-\pi x)}{\tan ^2 (\pi-\pi x)} . \frac{1}{1+ \cos (\pi-\pi x)} \)
    
    or \( \displaystyle \lim_{x \to 1} \frac{\sin^2 (\pi-\pi x)}{\tan ^2 (\pi-\pi x)} . \frac{1}{1+ \cos (\pi-\pi x)} \)
    
    or \( \left ( \displaystyle \lim_{x-1 \to 0} \cos (\pi-\pi x) \right )^2 . \left ( \displaystyle \lim_{x \to 1} \frac{1}{1+ \cos (\pi-\pi x)} \right ) \)
    
    or \( \left ( 1 \right )^2 . \left ( \frac{1}{1+ 1} \right ) \)
    
    or \( \frac{1}{2} \)
    
    This completes the solution

26. \( \displaystyle \lim_{x \to \theta} \frac{x \tan \theta -\theta \tan x}{x -\theta}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to \theta} \frac{x \tan \theta -\theta \tan x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{x \tan \theta - \theta \tan \theta + \theta \tan \theta - \theta \tan x}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \tan \theta + \theta (\tan \theta - \tan x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \frac{(x- \theta) \tan \theta} {x -\theta} + \displaystyle \lim_{x \to \theta} \frac{\theta (\tan \theta - \tan x)}{x -\theta} \)
    
    or \( \displaystyle \lim_{x \to \theta} \tan \theta - \displaystyle \lim_{x \to \theta} \frac{\theta (\tan x -\tan \theta )}{x -\theta} \)
    
    or \( \tan \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ (\tan x -\tan \theta )}{x -\theta} \)
    
    or \( \tan \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ \tan (x- \theta )(1+\tan x \tan \theta)}{x -\theta} \)
    
    or \( \tan \theta - \theta \displaystyle \lim_{x \to \theta} \frac{ \tan (x- \theta )}{x -\theta} \displaystyle \lim_{x \to \theta} (1+\tan x \tan \theta)\)
    
    or \( \tan \theta - \theta .1.(1+\tan^2 \theta)\)
    
    or \( \tan \theta - \theta \sec^2 \theta\)
    
    This completes the solution

27. \( \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\cos \theta -\sin \theta}{\theta -\frac{\pi}{4}}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\cos \theta -\sin \theta}{\theta -\frac{\pi}{4}} \)
    
    or \( \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{ \sqrt{2} (\cos \theta \frac{1}{\sqrt{2}}-\sin \theta \frac{1}{\sqrt{2}}) } {\theta -\frac{\pi}{4}} \)
    
    or \( \sqrt{2} \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\cos \theta \sin \frac{\pi}{4} -\sin \theta \cos \frac{\pi}{4}}{\theta -\frac{\pi}{4}}\)
    
    or \( - \sqrt{2} \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\sin \theta \cos \frac{\pi}{4} -\cos \theta \sin \frac{\pi}{4}}{\theta -\frac{\pi}{4}}\)
    
    or \( - \sqrt{2} \displaystyle \lim_{\theta \to \frac{\pi}{4}} \frac{\sin (\theta -\frac{\pi}{4}) }{\theta -\frac{\pi}{4}}\)
    
    or \( - \sqrt{2} .1\)
    
    or \( - \sqrt{2}\)
    
    This completes the solution

28. \( \displaystyle \lim_{x \to c} \frac{\sqrt{x}-\sqrt{c}}{\sin x-\sin c}\)
    

    Solution 👉 Click Here

    Given that
    \( \displaystyle \lim_{x \to c} \frac{\sqrt{x}-\sqrt{c}}{\sin x-\sin c} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{x-c}{(\sin x-\sin c)(\sqrt{x}+\sqrt{c})} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{x-c}{\sin x-\sin c} \frac{1}{\sqrt{x}+\sqrt{c}} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{x-c}{2 \cos (\frac{x+c}{2}) \sin (\frac{x-c}{2}) } \frac{1}{\sqrt{x}+\sqrt{c}} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{\frac{x-c}{2}}{ \sin (\frac{x-c}{2}) } \frac{1}{\cos (\frac{x+c}{2}) (\sqrt{x}+\sqrt{c})} \)
    
    or \( \displaystyle \lim_{x \to c} \frac{\frac{x-c}{2}}{ \sin (\frac{x-c}{2}) } \displaystyle \lim_{x \to c} \frac{1}{\cos (\frac{x+c}{2}) (\sqrt{x}+\sqrt{c})} \)
    
    or \( 1. \frac{1}{\cos c (\sqrt{c}+\sqrt{c})} \)
    
    or \( \frac{\sec c}{2\sqrt{c}} \)
    
    This completes the solution

---

29. Find the limits of
    1. \( \displaystyle \lim_{x \to 0} \frac{e^{6x}-1}{x}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{e^{6x}-1}{x} \)
       
       or \( \left ( \displaystyle \lim_{x \to 0} \frac{e^{6x}-1}{6x} \right ).6\) \( \because \displaystyle \lim_{x \to 0} \frac{e^{x}-1}{x}=1\)
       
       or \( 1.6\)
       
       or \( 6\)
       
       This completes the solution

    2. \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{x. 2^{x+1}}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{x. 2^{x+1}} \)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{x} . \frac{1}{2^{x+1}}\)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{2x} . \frac{1}{2^{x}}\)
       
       or \( \left ( \displaystyle \lim_{x \to 0} \frac{e^{2x}-1}{2x} \right ) . \left (\displaystyle \lim_{x \to 0} \frac{1}{2^{x}} \right ) \)
       
       or \( \left ( 1 \right ) . \left (\frac{1}{1} \right ) \) \( \because \displaystyle \lim_{x \to 0} \frac{e^{x}-1}{x}=1\)
       
       or \( 1\)
       
       This completes the solution

    3. \( \displaystyle \lim_{x \to 0} \frac{e^{ax}-e^{bx}}{x}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{e^{ax}-e^{bx}}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{e^{ax}-1-e^{bx}+1}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{(e^{ax}-1)-(e^{bx}-1)}{x} \)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{ax}-1}{x}- \displaystyle \lim_{x \to 0} \frac{e^{bx}-1}{x} \)
       
       or \( \displaystyle \lim_{x \to 0} \frac{e^{ax}}{ax} .a - \displaystyle \lim_{x \to 0} \frac{e^{bx}}{bx} .b \)
       
       or \( 1 .a - 1 .b \) \( \because \displaystyle \lim_{x \to 0} \frac{e^{x}-1}{x}=1\)
       
       or \( a - b \)
       
       This completes the solution

    ---

    4. \( \displaystyle \lim_{x \to 0} \frac{a^x+b^x-2}{x}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 0} \frac{a^x+b^x-2}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{a^x-1+b^x-1}{x} \)
       
       or\( \displaystyle \lim_{x \to 0} \frac{a^x-1}{x}+ \lim_{x \to 0} \frac{b^x-1}{x} \)
       
       or\( \log a+ \log b \) \( \because \displaystyle \lim_{x \to 0} \frac{a^x-1}{x}=\log a\)
       
       or\( \log (ab)\)
       
       This completes the solution

30. Evaluate the limits of
    1. \( \displaystyle \lim_{x \to 2} \frac{x-2}{\log (x-1)}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to 2} \frac{x-2}{\log (x-1)} \)
       
       or\( \displaystyle \lim_{x \to 2} \frac{x-2}{\log [1+(x-2)]} \)
       
       or\( \displaystyle \lim_{x \to 2} \frac{1}{\frac{\log [1+(x-2)]}{(x-2)}} \)
       
       or\( \frac{1}{ \displaystyle \lim_{x-2 \to 0} \frac{\log [1+(x-2)]}{(x-2)}} \)
       
       or\( \frac{1}{ 1} \) \( \because \displaystyle \lim_{x \to 0} \frac{\log (1+x)}{x}=1\)
       
       or\( 1 \)
       
       This completes the solution

    2. \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\log \left ( x- \frac{\pi}{2} +1\right )}\)
       

       Solution 👉 Click Here

       The limit takes the indeterminate form \(\frac{0}{0}\), we proceed further calculation
       Given that
       \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\log \left ( x- \frac{\pi}{2} +1\right )} \)
       
       or\( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin (\frac{\pi}{2}-x)}{\log \left [ 1+(x- \frac{\pi}{2}) \right ]} \)
       
       or\( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{- \sin (x-\frac{\pi}{2})}{\log \left [ 1+(x- \frac{\pi}{2}) \right ]} \)
       
       or\( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{- \sin (x-\frac{\pi}{2})}{(x-\frac{\pi}{2})} . \frac{x-\frac{\pi}{2}}{\log \left [ 1+(x- \frac{\pi}{2}) \right ]} \)
       
       or\( - \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{ \sin (x-\frac{\pi}{2})}{(x-\frac{\pi}{2})} . \frac{1}{ \displaystyle \lim_{x- \frac{\pi}{2} \to 0} \frac {\log \left [ 1+(x- \frac{\pi}{2}) \right ] }{ (x-\frac{\pi}{2})} } \)
       
       or\( - 1 . 1 \)
       
       or\( - 1 \)
       
       This completes the solution