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manionkr · Sep 3, 2008, 09:24 AM

I am trying to find the lim as x---> 0 for (sin5x)/(x^2)

I have come up with:

(5sin5x)/5x times (1/x) = (5)(1) x (1/x), but I can't figure out how to get the x out of the bottom to solve. I might have done the first part wrong.

ebaines · Sep 3, 2008, 12:09 PM

Try applying L'Hopital's rule: take the derivative with respect to x for both the numerator and denominator, and then find the limit as x goes to 0 for each, and divide:

$<br /> \displaystyle\lim_ {x \to 0} \frac {\sin(5x)} {x^2} = \displaystyle\lim_ {x \to 0} \frac { (\sin(5x))'} {(x^2)'} \\<br /> = \displaystyle\lim_ {x \to 0} \frac {\5 cos(5x)} {2x} \\<br />$

Can you take it from here for the last step to get the answer?

manionkr · Sep 3, 2008, 01:30 PM

Originally Posted by manionkr

I am trying to find the lim as x---> 0 for (sin5x)/(x^2)

I have come up with:

(5sin5x)/5x times (1/x) = (5)(1) x (1/x), but i can't figure out how to get the x out of the bottom to solve. i might have done the first part wrong.

to tell you the truth... no. I've never really been good with these types of limits. I don't even know what the next step would be. If that cos was a sin I would know how to do it, but its not.

ebaines · Sep 5, 2008, 05:45 AM

After having applied L'Hopital's rule, you have to evaluate the limit of 5cos(5x)/2x. The limit of cos(5x) as x goes to zero is 1, since cos(0)=1. The limit of 2x as x goes to zero is zero. So this problem becomes 5*1/0 = $\infty$ .