# 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 $f (x) = \{0, \infty \}$

#### Theorem on Limit of exponential function

1. Prove that $\displaystyle \lim_{x \to \infty} \left( 1+\frac{1}{x} \right )^x =e$
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
$\left( 1+\frac{1}{x} \right )^x =1+ x \frac{1}{x} +\frac{x(x-1)}{2!} \left( \frac{1}{x}\right)^2+\frac{x(x-1)(x-2)}{3!} \left( \frac{1}{x}\right)^3 +...$
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 $\displaystyle \lim_{x\to 0}\frac{e^x-1 }{x}=1$
Solution
$\displaystyle \lim_{x\to 0}\frac{e^x-1 }{x}$
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.