# Thread: Limit and Chain Rule Questions

1. ## Limit and Chain Rule Questions

Hi, can someone help me with this?

lim x->0 $\displaystyle \frac{x-tan(4x)}{sin(4x)}$

I tried multiplying by the conjugate then using the trig identity for tan, but I got a bunch of mess I don't know what to do with.

Also, for

lim x->0 $\displaystyle \frac{x^2-12x+32}{sin(x-4)}$

is the limit -8? just want to make sure that's right.

and how do I solve this:

let F(x) = h(g(f(x))), where f(4) = 7, g(7) = 8, f '(4) = 3, g '(7) = 8, and h '(8) = 2

then find F'(4)

confused as how to do this

thanks!

2. $\displaystyle \frac{x - \tan(4x)}{\sin(4x)} = \frac{x}{\sin(4x)} - \frac{\tan(4x)}{\sin(4x)} = \frac{4x}{4\sin(4x)} - \frac{\frac{\sin(4x)}{\cos(4x)}}{\sin(4x)} = \frac{1}{4} \cdot \frac{4x}{\sin(4x)} - \frac{1}{\cos(4x)}$
now take the limit as x approaches 0.

for the 2nd limit ... if x is approaching 0, then limit is $\displaystyle \frac{32}{\sin(-4)}$

last one, use the chain rule ...

$\displaystyle F'(x) = h'[g(f(x))] \cdot g'(f(x)) \cdot f'(x)$

now sub in x = 4 and evaluate.

3. thanks, still confused about the last one though

can anyone help me find F'(4) if
F(x) = h(g(f(x))), where f(4) = 7, g(7) = 8, f '(4) = 3, g '(7) = 8, and h '(8) = 2

not sure where to sub in 4 or whatever.

4. $\displaystyle F'(4) = h'[g(f(4))] \cdot g'(f(4)) \cdot f'(4)$

$\displaystyle F'(4) = h'[g(7)] \cdot g'(7) \cdot 3$

$\displaystyle F'(4) = h'[8] \cdot 8 \cdot 3$

$\displaystyle F'(4) = 2 \cdot 8 \cdot 3 = 48$

5. ahh of course, how did I not see that??

thanks!