L'Hôpital's rule

  • Thread starter LizzieL
  • Start date
  • #1
12
0

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?
 

Answers and Replies

  • #2
993
13
there is 1.92 missing in the numerator.
 

Related Threads on L'Hôpital's rule

  • Last Post
Replies
2
Views
886
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
12
Views
4K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
821
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
9
Views
2K
Top