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 \(2^x, 2^{-x} \)

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

Graph of two exponential function \(e^x, e^{-x} \)

Theorem on Limit of exponential function

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

  2. Prove that \(\displaystyle \lim_{x\to 0}\frac{e^x-1 }{x}=1 \)
    \(\displaystyle \lim_{x\to 0}\frac{e^x-1 }{x} \)
    or \(\displaystyle \lim_{x\to 0}\frac{\left ( 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right ) -1 }{x} \)
    or \(\displaystyle \lim_{x\to 0}\frac{\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+... }{x} \)
    or \(\displaystyle \lim_{x\to 0} \frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+... \)
    or \(\frac{1}{1!}+\frac{0}{2!}+\frac{0^2}{3!}+... \)
    or 1
    This completes the solution.

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