Basics

Calculus Level 2

Evaluate $\displaystyle \lim_{x \to 0} (\frac{e^{3x}-1}{x})$

Take $\displaystyle e =\lim_{x\to0} \left(1+\frac1x\right)^x$ .

1 solution

Akshat Sharda
Aug 22, 2015

$\Rightarrow \displaystyle \lim_{x \to 0 } (\frac{e^{3x}-1}{x})$

$\Rightarrow \displaystyle \lim_{ \color{#D61F06}{3x} \to 0 } [(\frac{e^{3x}-1}{\color{#D61F06}{3x}})×\color{#D61F06}{3}]$

When : $(x \to 0) \Rightarrow (\color{#D61F06}{3x}\to 0)$ .

$\Rightarrow \displaystyle \color{#D61F06}{3}×\lim_{y \to 0}(\frac{e^{y}-1}{y})$

Here : $y=\color{#D61F06}{3x}$ .

$\Rightarrow (\color{#D61F06}{3}×1) \Rightarrow \color{#D61F06}{\boxed{3}}$

cool solution man

Akash singh - 5 years, 9 months ago

$\huge ThAnKs!!$

Akshat Sharda - 5 years, 9 months ago

I suppose that it might be worth mentioning that since

$e^{y} = 1 + y + \dfrac{y^{2}}{2!} + \dfrac{y^{3}}{3!} + ....,$

we know that $e^{y} - 1 = y + O(y^{2}),$ and hence $\dfrac{e^{y} - 1}{y} = 1 + O(y),$

the limit of which as $y \rightarrow 0$ is $1.$

We could have also noted that $\dfrac{e^{3x} - 1}{x} = \dfrac{e^{x} - 1}{x} \times (e^{2x} + e^{x} + 1),$

and proceeded accordingly, or just directly used the expansion

$e^{3x} = 1 + 3x + \dfrac{(3x)^{2}}{2!} + \dfrac{(3x)^{3}}{3} + ....\Longrightarrow e^{3x} - 1 = 3x + O(x^{2}).$

Of course, there is always L'Hopital's rule, but that always seems like cheating when there are alternative ways of finding a limit. :)

Brian Charlesworth - 5 years, 9 months ago