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Limits of trigonometric functions

  • Thread starter jkeatin
  • Start date
  • #1
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Homework Statement



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

Homework Equations



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

The Attempt at a Solution



should i differentiate first? am i allowed to do that?
 
Last edited:

Answers and Replies

  • #2
66
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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
66
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does this all look correct?
 
  • #4
1,752
1
What book are you using? Chapter and problem number?
 
  • #5
66
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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
1,752
1
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
Ok then, you can help yourself! You're given a take home exam ... use your book.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
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First write the problem correctly. Is it
[tex]\lim_{x\rightarrow 2}\frac{tan^3(2+x)- tan(8)}{x}[/tex]
or is it
[tex]\lim_{x\rightarrow 2}\frac{tan((2+x)^3)- tan(8)}{x}[/tex]

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
66
0
the second one is right. but as x goes to 0
 
  • #9
66
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sec^2(x^3)(3x)
 
  • #10
66
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i think that is right and then i just substitute in 2 right? so its
6sec^2(8)
 

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