Question

What is `\lim _ { x \rightarrow 0 } \frac { e ^ { 2 x } - e ^ { x } } { x }`?

Answer 1

`lim_(x rarr 0) (e^(2x)-e^x)/(x) = 1`

Explanation:

Method 1 : Graphically
graph{(e^(2x)-e^x)/(x) [-8.594, 9.18, -1.39, 7.494]}

Although far from conclusive, it appears that:

`lim_(x rarr 0) (e^(2x)-e^x)/(x) = 1`

Method 2 : L'Hôpital's rule

The limit:

`lim_(x rarr 0) (e^(2x)-e^x)/(x)`

is of an indeterminate form `0/0`, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exits then:

`lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x))`

And so applying L'Hôpital's rule we get:

`lim_(x rarr 0) (e^(2x)-e^x)/(x) = lim_(x rarr 0) (2e^(2x)-e^x)/(1)`
`" "= 2-1`
`" "= 1`

Method 3 - Power Series

The power series for `e^x` is as follows;

`e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...`

And so we have:

`e^(2x) = 1 + 2x + (2x)^2/(2!) + (2x)^3/(3!) + (2x)^4/(4!) + ...`

Therefore;

`(e^(2x)-e^x)/(x) = { (1 + 2x + O(x^2)) - (1 + x + O(x^2)) } / x`
`" " = { 1 + 2x + O(x^2) - 1 - x + O(x^2) } / x`
`" " = { x + O(x^2) } / x`
`" " = 1 + O(x)`

And so:

`lim_(x rarr 0) (e^(2x)-e^x)/(x) = lim_(x rarr 0) (1 + O(x))`
`" " = 1`