Limit of a Trigonometric Function

  • Thread starter dekoi
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  • #1
dekoi
Question:

lim(x->0) for (tanx - sinx) / (sinx)^2

This is what I got:

= (sinx-sinxcosx) / (cosx)(sinx)^2
= (sinx)(1-cosx) / (sinx)(sinx)(cosx)
= (1 - cosx) / (sinx)(cosx)

However, I can't figure out what to do from this step, as the limit still equals 0/0 at this stage.
 

Answers and Replies

  • #2
mezarashi
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A 0/0 answer is the right prerequisite for using L'Hopital's rule.

That is, the limit of any such function f(x) = h(x)/g(x), is h'(x)/g'(x).
Try differentiating the top and bottom separately and plug in the numbers again.
 
  • #3
dekoi
I haven't learned that yet.

We are expected to solve the problem with using only limit laws and the fact that the lim (x->0) for sinx / x equals 1.
 
  • #4
dekoi
??? .
 
  • #5
mezarashi
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Then you need to approach the problem differently. It is purely an algebriac/trigonometric problem. The strategy is to rid the denominator of any possible 0 terms (i.e. sin x).

Edit - Here, try this:
[tex] \frac{\tan x - \sin x}{sin^2x}[/tex]

[tex]\frac{\frac{\sin x - \sin x \cos x}{\cos x}}{1 - \cos^2 x}[/tex]

Keep in mind:
[tex](1-a^2) = (1-a)(1+a)[/tex]
 
Last edited:
  • #6
HallsofIvy
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Do you know that [tex]lim_{x\rightarrow0}\frac{sin x}{x}= 1[/tex]?
Do you know that [tex]lim_{x\rightarrow0}\frac{1- cos x}{x}= 0[/tex]?

Can you figure out how to write [tex]\frac{1-cos x}{sin x}[/tex] in terms of [tex]\frac{sin x}{x}[/tex] and [tex]\frac{1- cos x}{x}[/tex]?
 
  • #7
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(1 - cosx) / (sinx)(cosx) *
(1+cosx) / (1+cosx) = ...

Or:
(tanx - sinx) / (sinx)^2 =
tanx (1 - cosx) / (1 - (cosx)^2) = ...
(look up Mezarashi's hint)
 
Last edited:
  • #8
lim x->0 of (tanx - sinx)/(sinx)^2

lim x->0 of (tanx)/(sinx)^2 - (sinx)/(sinx)^2

lim x->0 of (sinx)/(cosx(sinx)^2) - (1/sinx)

lim x->0 (1/cosxsinx) - (1/sinx)

lim x->0 (1-cosx)/(cosxsinx)

lim x->0 (-(cosx-1)/x) / (cosxsinx)/(x))

lim x->0 (-0)/ (1(1)) = 0

so the final answer is 0
 

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