# Limits of trigonometric functions

1. Jul 6, 2008

### jkeatin

1. The problem statement, all variables and given/known data

lim x --->0 [tan(2+x)^3 - tan8]/x

2. Relevant equations

f'(a)=lim h-->0 f(a+h)-f(a)/h

3. The attempt at a solution

should i differentiate first? am i allowed to do that?

Last edited: Jul 6, 2008
2. Jul 6, 2008

### jkeatin

f(a)=tan8
f(a+h)=tan(x+2)^3
h= x
a+h=2+x=2+h so a= 2
f(a)=f(2)=tan8
f(x)=tanx^3
f(2)=tan2^3=tan8
f'(x)=secx^2 * 3x
f'(2)= sec2^2 * 6= 6sec4

3. Jul 6, 2008

### jkeatin

does this all look correct?

4. Jul 6, 2008

### rocomath

What book are you using? Chapter and problem number?

5. Jul 6, 2008

### jkeatin

it a problem for a test, i just need to evaluate the limit above, this is what the teacher showed to do for a similar problem

6. Jul 6, 2008

7. Jul 6, 2008

### HallsofIvy

First write the problem correctly. Is it
$$\lim_{x\rightarrow 2}\frac{tan^3(2+x)- tan(8)}{x}$$
or is it
$$\lim_{x\rightarrow 2}\frac{tan((2+x)^3)- tan(8)}{x}$$

If it is the second then it can be interpreted as the derivatve of tan(x3) at x= 2. What is that derivative? (it is NOT sec2(x^2) (3x)[. Use the chain rule.)

8. Jul 6, 2008

### jkeatin

the second one is right. but as x goes to 0

9. Jul 6, 2008

### jkeatin

sec^2(x^3)(3x)

10. Jul 6, 2008

### jkeatin

i think that is right and then i just substitute in 2 right? so its
6sec^2(8)