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## Homework Statement

I have

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

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":

[tex] L = \lim_{x\rightarrow 0} \Big( {\frac{(-sin 1.92x)1.92} {2.33e^{2.33x}-2.33}}\Big)[/tex]

Getting rid of "1.92" gives

[tex] \lim_{x\rightarrow 0} \Big( {\frac{-sin1.92x} {1.21e^{2.33x}-1.21}}\Big) [/tex]

Then since [tex]{\frac {0} {0}}[/tex], I'll use the rule once more:

[tex] \lim_{x\rightarrow 0} {\frac{-cos 1.92x} {2.82e^{2.33x}}} = -0.35 [/tex]

What am I doing wrong?