# Question #67981

Apr 21, 2017

${\lim}_{x \to 0} \frac{{e}^{3 x} - {e}^{x}}{x} = 2$
${\lim}_{x \to 0} \frac{{e}^{3 x} - e x}{x} = \infty$

#### Explanation:

If you didn't make a typo and you meant ${\lim}_{x \to 0} \frac{{e}^{3 x} - e x}{x}$ Then there is nothing else to do but substitute: $\frac{1 - 0}{0} \to \infty$

If you meant ${\lim}_{x \to 0} \frac{{e}^{3 x} - {e}^{x}}{x}$ Then we have an indetermination $\frac{1 - 1}{0} = \frac{0}{0}$ And we can use Bernoulli l'Hospital:

${\lim}_{x \to 0} \frac{{e}^{3 x} - {e}^{x}}{x} = {\lim}_{x \to 0} \frac{\left(\frac{d}{\mathrm{dx}}\right) \left({e}^{3 x} - {e}^{x}\right)}{\left(\frac{d}{\mathrm{dx}}\right) x} = {\lim}_{x \to 0} \frac{3 {e}^{3 x} - {e}^{x}}{1} = 2$

Apr 21, 2017

$2.$

#### Explanation:

We use this Standard Form of Limit : ${\lim}_{x \to 0} \frac{{e}^{x} - 1}{x} = 1.$

This will help us evaluate the reqd. limit without L'Hospital's Rule.

Reqd. Limit$= {\lim}_{x \to 0} \frac{{e}^{3 x} - {e}^{x}}{x}$

$= {\lim}_{x \to 0} \left\{\frac{\left({e}^{x}\right) \left({e}^{2 x} - 1\right)}{x}\right\}$

$= \left\{{\lim}_{x \to 0} {e}^{x}\right\} \left[2 \left\{{\lim}_{\left(2 x\right) \to 0} \frac{{e}^{2 x} - 1}{2 x}\right\}\right]$

$= \left({e}^{0}\right) \left[2 \left\{1\right\}\right]$

$\therefore \text{ The Reqd. Lim.=} 2.$

Enjoy Maths.!