Question 9130a

Dec 15, 2017

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

Explanation:

${\lim}_{x \to 0} \frac{\left(3 \left({e}^{x} - 1\right)\right) - 3 x}{x \left({e}^{x} - 1\right)}$

If we plug $0$ directly into our equation, we get the outcome:

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

(Quotations to ensure I'm not stating that the two are equal and is actually undefined)

However, this allows us to use L'Hospital's Rule.

We need to have $\frac{0}{0}$ or $\frac{\infty}{\infty}$ for L'Hospital's Rule to be applicable.

Thus, we can move forward with that rule, but lets first distribute to help up with taking derivatives:

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

=>lim_(x->0)(d/dx[3e^x-3x-3])/(d/dx[xe^x-x]) = lim_(x->0)(3e^x-3)/(e^x+xe^x-1#

Now, let's try plugging in $0$ for $x$:

${\lim}_{x \to 0} \frac{3 {e}^{x} - 3}{{e}^{x} + x {e}^{x} - 1} = \frac{3 {e}^{0} - 3}{{e}^{0} - 0 {e}^{0} - 1}$"="$\frac{0}{0}$

The answer is still undefined, but we can apply L'Hospital's Rule once again since we got an output of $\frac{0}{0}$.

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

$= {\lim}_{x \to 0} \frac{3 {e}^{x}}{2 {e}^{x} + x {e}^{x}} = \frac{3 {e}^{0}}{{e}^{0} + 0 {e}^{0} - 1} = \frac{3}{2}$

Thus,

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