Question about this problem using L'Hospital's rule

lim x->0 6x-sin(6x)/6x-tan(6x) I get that it is 0/0 so I use the rule. I get 6-6cos(6x)/6-6sec^2(6x) and you get 0/0 and you use the rule again. So you get

36sin(6x)/-36sec^2(6x)tan(6x) which is 0/0 so you use the rule again. I'm confused on the derivative of -36sec^2(6x)tan(6x). The answer comes out to -1/2 but I'm not sure how. Am I even going down the right path?

Re: Question about this problem using L'Hospital's rule

Quote:

Originally Posted by

**Alucard2487** lim x->0 6x-sin(6x)/6x-tan(6x) I get that it is 0/0 so I use the rule. I get 6-6cos(6x)/6-6sec^2(6x) and you get 0/0 and you use the rule again. So you get

36sin(6x)/-36sec^2(6x)tan(6x) which is 0/0 so you use the rule again. I'm confused on the derivative of -36sec^2(6x)tan(6x). The answer comes out to -1/2 but I'm not sure how. Am I even going down the right path?

You missed a factor of 2 when you took the last derivative of the denominator, should look like this

$\displaystyle \lim_{x \to 0} \ \frac{36sin(6x)}{-36 \cdot 2 sec(6x)sec(6x)tan(6x)} \\ \\ \text{At this point simplify as much as you can by converting secx and tanx into sine and cosines. It will look like} \\ \\ \lim_{x \to 0} \frac{cos^3(6x)}{-2} \\ \\ \text{ direct substitution now gives -1/2} $

:)

Re: Question about this problem using L'Hospital's rule

It's always something small. I messed up the chain rule. Thank you for your help.