Introduction to Limit

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Introduction

In Mathematics, limit is defined as a value that a function approaches for the given point. It is always concerns about the behavior of a function at a particular point.

1. Mathematics मा Limit भनेको फलनको मान हो। जुन मान given point मा कन्सन्ट्रेट भएको हुन्छ। 
2. गणितज्ञ Henkel(१८७१) का अनुसार आजभोलि हामीले कुरा गर्ने limit को परिभाषाको Euclid को Elements मा भएको बुक 10 को Proposotion 1 मा आधारित छ जसमा Euclid र Archimedes ले चर्चा गर्ने गरेको Method of exhaustion को प्रयोग भएको छ।
3. Calculus मा continuity, derivatives and integrals जस्ता कुराहरुको चर्चा गर्न limit को अध्य्यन जरुरी छ।
4. limit को परिभाषामा प्रयोग गर्ने ε-δ notation को चर्चा १९औ शताब्दीबाट Bernard Bolzano ले शुरु गरेको पाइन्छ, औपचारिक रुपमा 1821 मा Augustin-Louis Cauchy र पछि Karl Weierstrass ले limit को परिभाषा मा ε-δ notation को प्रयोग सुरु गरेको पाईन्छ।
5. limit को ठिक तल प्रयोग गर्ने arrow चिन्हको प्रयोग २०औँ शताब्दीमा G. H. Hardy ले उनको किताब a course of pure mathematics बाट सुरु गरेको पाइन्छ ।

Meaning of $x\to a$

Consider a function
$f(x)=\frac{x^2-1}{x-1}$
We know that function is NOT defined at x=1.
However, what happens to f(x) near the value x=1?

0,0

If we substitute small values for x near to 1, then we find the value of f(x) near to 2
This is shown in a table below.

	x<1					x=1	x>1
x	0.5	0.6	0.7	0.8	0.9	1	1.1	1.2	1.3	1.4	1.5
f(x)	1.5	1.6	1.7	1.8	1.9	2	2.1	2.2	2.3	2.4	2.5

Tha table value shos that, the closer that x gets to 1, the closer the value of the function f(x) to 2.
In such cases, we call it
f(x)=2 as x tends to 1

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Intuitive Definition of Limit

Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
Then we say that
${\displaystyle \underset{x\to a}{lim}f(x)=L}$
if for a given number $ϵ>0$, there exists a number $\delta >0$such that
|f(x)-L|<$ϵ$ whenever |x-a|<$\delta$

In this definition

1. Given a number a, we choose δ-neighbourhood of a, δ is positive AND can be small enough as we like such that |x - a|< δ
2. Now, we will try to find, ε-neighbourhood of a given number L, ε is positive AND can be small enough as satisfied such that |f(x) - L|< ε
3. If a small change in δ implies a small change ε, then the limit exists at a.
4. If a small change in δ implies a LARGE change ε, or vice versa, then the limit does NOT exist at a.

Example: Limit exist at x=a

0,0

Show δ nbd

Show ε nbd

a

δ = 0.50

L

ε = 2.641

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How to use the applet

1. Click on the "Show δ-neighbourhood of 'a'" check box.
2. Given a number a, adjust δ-neighbourhood of a (drag the point a or the slider δ) , so that |x - a|< δ , where x is any point inside the δ-neighbourhood
3. Now, Click on the "Show ε-neighbourhood of 'L'" check box.
4. Try to find ε-neighbourhood of L (largest distance from L) , such that |f(x) - L|< ε , where f(x) is any point inside the ε-neighbourhood and the result is valid for all |x - a|< δ
   
   सबै |x - a|< δ को लागी |f(x) - L|< ε  हुने गरि ε-zone बानउन सकिन्छ भने limit exist हुन्छ ।
   
   Once a ε is found, any higher ε is always accepted.
   Once a δ is satisfied, any smaller δ is always accepted.

More Explanation

The intuitive definition says that

1. determine a number δ>0
2. take any x in the region, i.e. between a+δ and a−δ, then this x will be closer to a, that is |x-a|<δ
3. identify the point on the graph that our choice of x gives, then this point on the graph will lie in the intersection of the ε region. This means that this function value f(x) will be closer to L , that is |f(x)-L|<ε
   Means
   if we take any value of x in the δ region then the graph for those values of x will lie in the ε region.
4. Once a δ is found, any smaller delta is acceptable, so there are an infinite number of possible δ's that we can choose.
5. the function has limit at given x

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Example: Limit Does not exist at x=0

0,0

Show δ nbd

Show ε nbd

a

δ = 0.50

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Empirical Definition of Limit

Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
Then we say that ${\displaystyle \underset{x\to a}{lim}f(x)=L}$ if
all of the following three conditions hold

1. ${\displaystyle \underset{x\to a^{-}}{lim}f(x)}$ exists=LHS
2. ${\displaystyle \underset{x\to a^{+}}{lim}f(x)}$ exists=RHS
3. LHS=RHS

For Example

1. In Figure (1), We see that the graph of f(x) has a hole at a. In fact, f(a) is undefined.[But, Limit exists at x=2]
2. In Figure (2), f(a) is defined, but the function has a jump at a.[So, Limit does NOT exist at x=2]
3. In Figure (3), f(a) is defined, but the function has a gap at a.[But, Limit exists at x=2]

0,0	0,0	0,0
$f(x)=\frac{x^2-1}{x-1}$	$f(x)=\{\begin{array}{ll}x+1& \text{for }x\le 2\\ x+2& \text{for }x>2\end{array}$	$f(x)=\{\begin{array}{ll}x+1& \text{for }x\ne 2\\ 4& \text{for }x=2\end{array}$

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Example 1

What do you mean by the left hand limit and right hand limit of a function? What is the condition for the limit of a function to exist at a point?
Prove that ${\displaystyle \underset{x\to 0}{lim}|x|=0}$ but ${\displaystyle \underset{x\to 0}{lim}\frac{|x|}{x}}$ does not exist.

Solution 👉 Click Here

Left-hand limit

The left-hand limit of a function f(x) at a particular point x=a is used to determine the behavior of the function f(x) as it approaches a from the left. Left-Hand Limit (LHL): is defined as
$LHL={\displaystyle \underset{x\to a^{-}}{lim}f(x)}$

This limit represents the behavior of the function f(x) as x gets closer to a while staying to the left of a.

0,0

–o+←↓↑→

f(x)

0.00

Right-hand limit

The right-hand limit of a function f(x) at a particular point x=a is used to determine the behavior of the function f(x) as it approaches a from the right. Right-Hand Limit (RHL): is defined as
$RHL={\displaystyle \underset{x\to a^{+}}{lim}f(x)}$

This limit represents the behavior of the function f(x) as x gets closer to a while staying to the right of a.

0,0

–o+←↓↑→

f(x)

0.00

Condition for the limit of a function to exist at a point

let f(x) be a function, then condition for the limit of a function f(x) to exist at a point a is that
$LHL=RHL$

or ${\displaystyle \underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)}$

This limit represents the behavior of the function f(x) as x gets closer to the point a, from both sides:left and right.

${\displaystyle \underset{x\to 0}{lim}|x|=0}$

The left hand limit is
$LHL={\displaystyle \underset{x\to 0^{-}}{lim}|x|=\underset{x\to 0^{-}}{lim}-x=}$ 0

The right hand limit is
$RHL={\displaystyle \underset{x\to 0^{+}}{lim}|x|=\underset{x\to 0^{+}}{lim}x=}$ 0

Since,
LHL=RHL

The limit exist at x=0, and the limit value is 0

${\displaystyle \underset{x\to 0}{lim}\frac{|x|}{x}}$ does NOT exist

The left hand limit is
$LHL={\displaystyle \underset{x\to 0^{-}}{lim}\frac{|x|}{x}=\underset{x\to 0^{-}}{lim}\frac{-x}{x}=}$ -1

The right hand limit is
$RHL={\displaystyle \underset{x\to 0^{+}}{lim}\frac{|x|}{x}=\underset{x\to 0^{+}}{lim}\frac{x}{x}=}$ 1

Since,
LHL≠RHL

The limit does NOT exist at x=0

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Example 2

Find the limit if exist for a function $f(x)=\frac{1}{1-x}$ at x=1

Solution 👉 Click Here

Given that
$f(x)=\frac{1}{1-x}$

At x=1, we compute the following
The functional value is
$f(x)=\frac{1}{1-x}=$ ∞

The left hand limit is
$LHL={\displaystyle \underset{x\to 1^{-}}{lim}f(x)=\underset{h\to 0^{-}}{lim}\frac{1}{1-(1-h)}=\underset{h\to 0}{lim}\frac{1}{h}=}$ ∞

The right hand limit is
$RHL={\displaystyle \underset{x\to 1^{+}}{lim}f(x)=\underset{h\to 0^{+}}{lim}\frac{1}{1-(1+h)}=\underset{h\to 0}{lim}\frac{1}{-h}=}$ -∞

Since,
LHL ,RHL,f(x) are NOT equal

The function is NOT continuous at x=1

0,0

f(x)

This completes the solution

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Indeterminate Form

The term "indeterminate" in mathematics refers to a situation where the value of an expression cannot be determined or uniquely identified based solely on its form or appearance.

1. $\frac{0}{0}$
   

   In the case of the expression "$\frac{0}{0}$" it is called indeterminate because it doesn't provide enough information to definitively determine the value of the expression.

   For example

   $\frac{1}{1}=1$	$\frac{1}{0}=\mathrm{\infty }$	$\frac{0}{1}=0$
   $\frac{2}{2}=1$	$\frac{2}{0}=\mathrm{\infty }$	$\frac{0}{2}=0$
   $\frac{3}{3}=1$	$\frac{3}{0}=\mathrm{\infty }$	$\frac{0}{3}=0$
   …	…	…
   $\frac{a}{a}=1$	$\frac{a}{0}=\mathrm{\infty }$	$\frac{0}{a}=0$
   $\frac{0}{0}=1$	$\frac{0}{0}=\mathrm{\infty }$	$\frac{0}{0}=0$

   Here, $\frac{0}{0}$ creates a situation where there is uncertainty about how the fraction $\frac{0}{0}$ as a whole behaves. In other words, knowing that both the numerator and denominator are approaching zero doesn't immediately mean $\frac{0}{0}$ will approach a specific finite value, approach infinity, or approach zero. The behavior of the fraction depends on the specific functions involved and how they approach zero.

2. $\frac{\mathrm{\infty }}{\mathrm{\infty }}$
   Usually $\frac{\mathrm{\infty }}{number}=\mathrm{\infty }$ and $\frac{number}{\mathrm{\infty }}=0$. So the top pulls the limit up to infinity and the bottom tries to pull it down to 0. So who wins?
3. $0.\mathrm{\infty }$
   Usually 0 · (number) = 0 and (number) · ∞ = ∞. So one piece tries to pull the limit down to zero, and the other tries to pull it up to ∞. Does one side win?
4. $\mathrm{\infty }-\mathrm{\infty }$
   In general ∞ − (number) = ∞, but (number) − ∞ = −∞. So who wins?
5. ${\mathrm{\infty }}^0$
   In general ∞ raised to any positive power should be equal to ∞, ∞ raised to a negative power is 0, and anything raised to the zero should be equal to 1. So who wins?
6. $1^{\mathrm{\infty }}$
   Usually 1 raised to any power is just equal to 1. But fractions raised to the ∞ goes to zero, and numbers larger than 1 raised to the ∞ should go off to ∞. So where does $1^{\mathrm{\infty }}$ go?
7. $0^0$
   In general zero raised to any positive power is just zero, but but anything raised to the zero should be equal to 1. So which is it?

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Limit of algrabic function

1. ${\displaystyle \underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}}$
   

   Solution
   We know that
   $\frac{x^n-a^n}{x-a}=\frac{(x-a)(x^{n-1}+a{x}^{n-2}+a^2{x}^{n-3}+...+a^{n-1})}{x-a}$
   or $\frac{x^n-a^n}{x-a}=(x^{n-1}+a{x}^{n-2}+a^2{x}^{n-3}+...+a^{n-1})$
   Thus, taking limit as $x\to a$, we get
   ${\displaystyle \underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=\underset{x\to a}{lim}(x^{n-1}+a{x}^{n-2}+a^2{x}^{n-3}+...+a^{n-1})}$
   or ${\displaystyle \underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=(a^{n-1}+a.a^{n-2}+a^2.a^{n-3}+...+a^{n-1})}$
   or ${\displaystyle \underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}}$
   This completes the proof

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Limit of trigonometric function

Trigonometry is branch of mathematics that deals about Triangle. The trigonometric ratio with reference to an angle x is called trigonometric function. For example,
f(x)= sinx

In this section we learn about two very specific but important trigonometric limits, and how to use them; and other tricks to find most other limits of trigonometric functions. The first involves the sine function, and the limit is
${\displaystyle \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1}$

0,0

Here's a graph of $f(x)=\frac{\sin x}{x}$, showing that it has a hole at x = 0. Our task in this section will be to prove that the limit from both sides of this function is 1.

Theorems on Limit of trigonometric function

Area of triangle $OPM=\frac{1}{2}\sin \theta \cos \theta$

त्रिभुजको क्षेत्रफल $\frac{1}{2}\text{ base}\times \text{height}$ हुने भएकोले $\mathrm{△}OPM=\frac{1}{2}\sin \theta \cos \theta$ हुन्छ ।

0,0

O

P(x,y)

sinθ

M

cosθ

A

θ

Area of sector $OPA=\frac{1}{2}\theta$

वृतको चाँदक्षेत्रको क्षेत्रफल $\frac{1}{2}{r}^2\theta$ हुने भएकोले sector $OPA=\frac{1}{2}\theta$ हुन्छ ।

0,0

O

P

M

A

θ

Area of triangle $OBA=\frac{1}{2}\tan \theta$

त्रिभुजको क्षेत्रफल $\frac{1}{2}\text{ base}\times \text{height}$ हुने भएकोले $\mathrm{△}OBA=\frac{1}{2}\tan \theta$ हुन्छ ।

0,0

O

P

M

A

B

θ

In the figure above
Area of triangle OMP=$\frac{1}{2}\sin \theta \cos \theta$

Area of sector OAP=$\frac{1}{2}\theta$

Area of triangle OAB=$\frac{1}{2}\tan \theta$

Now
Area of triangle OMP $\le$ Area of sector OAP $\le$ Area of triangle OAB

or $\frac{1}{2}\sin \theta \cos \theta \le \frac{1}{2}\theta \le \frac{1}{2}\tan \theta$

or $\sin \theta \cos \theta \le \theta \le \tan \theta$

or $\cos \theta \le \frac{\theta }{\sin \theta }\le \frac{1}{\cos \theta }$

or $\frac{1}{\cos \theta }\ge \frac{\sin \theta }{\theta }\ge \cos \theta$

Taking limit as $\theta \to 0$, we get

${\displaystyle \underset{\theta \to 0}{lim}\frac{1}{\cos \theta }\ge \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }\ge \underset{\theta \to 0}{lim}\cos \theta }$

or ${\displaystyle \frac{1}{\cos 0}\ge \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }\ge \cos 0}$

or ${\displaystyle \frac{1}{1}\ge \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }\ge 1}$
or ${\displaystyle 1\ge \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }\ge 1}$

or ${\displaystyle \underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1}$
This completes the proof.

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More Theorems on Limit of trigonometric function

1. The another important limit involves the cosine function, specifically the function
   ${\displaystyle \underset{\theta \to 0}{lim}\frac{\cos \theta -1}{\theta }=0}$

   0,0

   Here's a graph of $f(x)=\frac{\cos x-1}{x}$, showing that it has a hole at x = 0. Our task in this section will be to prove that the limit from both sides of this function is 0.

   Prove that ${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos x}{x}=0}$
   Solution
   The limit is
   ${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos x}{x}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos x}{x}\times \frac{1+\cos x}{1+\cos x}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{1-{\cos }^{2}x}{x(1+\cos x)}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x(1+\cos x)}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{\sin x}{x}\times \underset{x\to 0}{lim}\frac{\sin x}{1+\cos x}}$
   or $1\times \frac{0}{1+1}$
   or 0
   This completes the proof

2. ${\displaystyle \underset{x\to 0}{lim}\sin x=0}$
3. ${\displaystyle \underset{x\to 0}{lim}\cos x=1}$
4. ${\displaystyle \underset{x\to 0}{lim}\frac{\tan x}{x}=1}$

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Limit of Exponential function

A function of the form $f(x)=a^x$ where base ‘a’ is constant (a>0) and the exponent ‘x’ is variable, is called exponential function.

For example,
$f(x)=2^x$
is an exponential function.

Graph of two exponential function $2^x,2^{-x}$

0,0

f(x)=2^x

f(x)=2^-x

The great Swiss mathematician Leonhard Euler (1707-1783) has introduced the number e (e = 2.7182818284….). This value e is useful to define exponential function.
The function $f(x)=e^x$ is called standard exponential function.
In this definition of $f(x)=e^x$

1. Domain of $f(x)=\{-\mathrm{\infty },\mathrm{\infty }\}$
2. Range of $f(x)=\{0,\mathrm{\infty }\}$

Graph of two exponential function $e^x,e^{-x}$

0,0

f(x)=e^x

f(x)=e^-x

Theorem on Limit of exponential function

1. Prove that ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=e}$
   Solution
   We know that
   1. $e=1+1!+\frac{1}{2!}+\frac{1}{3!}+...$
   2. $(1+x{)}^n=1+nx+\frac{n)n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...$
   Thus
   ${(1+\frac{1}{x})}^x=1+x\frac{1}{x}+\frac{x(x-1)}{2!}{\left(\frac{1}{x}\right)}^2+\frac{x(x-1)(x-2)}{3!}{\left(\frac{1}{x}\right)}^3+...$
   or ${(1+\frac{1}{x})}^x=1+1+\frac{1(1-\frac{1}{x})}{2!}+\frac{1(1-\frac{1}{x})(1-\frac{2}{x})}{3!}+...$
   Taking limit as $x\to \mathrm{\infty }$, we get
   ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=1+1+\underset{x\to \mathrm{\infty }}{lim}\frac{1(1-\frac{1}{x})}{2!}+\underset{x\to \mathrm{\infty }}{lim}\frac{1(1-\frac{1}{x})(1-\frac{2}{x})}{3!}+...}$
   or ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=1+1+\frac{1(1-0)}{2!}+\frac{1(1-0)(1-0)}{3!}+...}$
   or ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=1+1+\frac{1}{2!}+\frac{1}{3!}+...}$
   or ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=e}$
   This completes the Proof.

2. Prove that ${\displaystyle \underset{x\to 0}{lim}\frac{e^x-1}{x}=1}$
   Solution
   ${\displaystyle \underset{x\to 0}{lim}\frac{e^x-1}{x}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...)-1}{x}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...}{x}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+...}$
   or $\frac{1}{1!}+\frac{0}{2!}+\frac{0^2}{3!}+...$
   or 1
   This completes the solution.

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Limit of logarithmic function

A logarithm function is an exponent of exponential function. For example,
if $a^x=y$, then $x={\log }_{a}y$.
In this definition
Log is the exponent, (or, exponent= Log)
if
$3^2=9$ then $2={\log }_{3}9$

In general, a function of the form $f(x)={\log }_{e}x$ called logarithmic function.
where

• Domain of f (x) = $(0,\mathrm{\infty })$
• Range of f (x) =$(-\mathrm{\infty },\mathrm{\infty })$

Properties of logarithmic function

1. Product property: $\log a(x.y)={\log }_{a}x+\log a_y$
2. Quotient property: ${\log }_a(x/y)={\log }_{a}x-{\log }_{a}y$
3. Power property: ${\log }_a{x}^{n}n=n{\log }_{a}x$
4. ${\log }_{a}a=1,{\log }_{a}1=0$
5. ${\log }_{a}m={\log }_{a}b\times {\log }_{b}m$

Graph of exponential and logarithem function

Log is the reflection of exponential function about y=x line, which is shown in a graph given below

0,0

f(x)=log(x)

f(x)=e^x

y=x

Theorems on Limit of logarithmic function

1. ${\displaystyle \underset{x\to 0}{lim}\frac{{\log }_e(1+x)}{x}=1}$
2. ${\displaystyle \underset{x\to 0}{lim}\frac{{\log }_e(1-x)}{-x}=1}$
3. ${\displaystyle \underset{x\to 0}{lim}\frac{a^x-1}{x}=\log a}$
   We know that
   $\frac{d}{dx}{a}^x=\frac{d}{dx}{e}^{\log (a^x)}$
   or $\frac{d}{dx}{a}^x=\frac{d}{dx}{e}^{x\log a}$
   or $\frac{d}{dx}{a}^x=\log a.e^{x\log a}$
   or $\frac{d}{dx}{a}^x=\log a.e^{\log a^x}$
   or $\frac{d}{dx}{a}^x=\log a.a^x$
   Thus, the limit is
   ${\displaystyle \underset{x\to 0}{lim}\frac{a^x-1}{x}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{\frac{d}{dx}(a^x-1)}{\frac{d}{dx}(x)}}$
   or ${\displaystyle \underset{x\to 0}{lim}\frac{\log a.a^x}{1}}$
   or $\log a.a^0$
   or $\log a$
   This completes the proof

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