# How do you find lim t^2/(e^t-t-1) as t->0 using l'Hospital's Rule?

##### 1 Answer
Feb 27, 2017

The limit has value $2$.

#### Explanation:

L'Hospital's Rule states for ${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ if and only if $f \left(a\right) = g \left(a\right) = 0$

Checking, the limit is currently of the form $\frac{0}{0}$, so we can use l'Hospital's Rule.

The derivative of ${t}^{2}$ is $2 t$. The derivative of ${e}^{t} - t - 1$ is ${e}^{t} - 1$. The limit becomes

${\lim}_{t \to 0} \frac{2 t}{{e}^{t} - 1}$

$\frac{2 \left(0\right)}{{e}^{0} - 1}$

$\frac{0}{0}$

We are in the same position as we were originally. We will once again apply l'Hospitals. The derivative of $2 t$ is $2$ and the derivative of ${e}^{t} - 1$ is ${e}^{t}$. The limit becomes

${\lim}_{t \to 0} \frac{2}{e} ^ t$

$\frac{2}{e} ^ 0$

$\frac{2}{1}$

$2$

A graphical verification yields the same result.

graph{x^2/(e^x - x - 1) [-10, 10, -5, 5]}

Hopefully this helps!