Question

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

Answer 1

The limit has value `2`.

Explanation:

L'Hospital's Rule states for `lim_(x->a) (f(x))/(g(x)) = lim_(x-> a) (f'(x))/(g'(x))` if and only if `f(a) = g(a) =0`

Checking, the limit is currently of the form `0/0`, so we can use l'Hospital's Rule.

The derivative of `t^2` is `2t`. The derivative of `e^t - t - 1` is `e^t - 1`. The limit becomes

`lim_(t->0) (2t)/(e^t - 1)`

`(2(0))/(e^0 - 1)`

`0/0`

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

`lim_(t->0) 2/e^t`

`2/e^0`

`2/1`

`2`

A graphical verification yields the same result.

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

Hopefully this helps!