# L'Hôpital's rule

## Homework Statement

I have

$$L = \lim_{x\rightarrow 0} \Big( {\frac{\cos(1.92x)-1} {e^{2.33x} - 1 -2.33x}} \Big)$$

I'm meant to use L'Hôpital's rule finding the limit, maybe twice.

## The Attempt at a Solution

So, there's clearly something I have misunderstood, and hoping you might tell me what it is.

"Solution":
$$L = \lim_{x\rightarrow 0} \Big( {\frac{(-sin 1.92x)1.92} {2.33e^{2.33x}-2.33}}\Big)$$

Getting rid of "1.92" gives

$$\lim_{x\rightarrow 0} \Big( {\frac{-sin1.92x} {1.21e^{2.33x}-1.21}}\Big)$$
Then since $${\frac {0} {0}}$$, I'll use the rule once more:

$$\lim_{x\rightarrow 0} {\frac{-cos 1.92x} {2.82e^{2.33x}}} = -0.35$$
What am I doing wrong?