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Trig Limits

  • #1
i understand the basic lim x->0 (sinx/x) = 1 thimg, but if its something more advanced im clueless, such as:

lim 3x^2/(1-cosx) where x->0

how would you solve this one, or something like lim sin(cosx)/cosx where x-> pi/2?

thanks for any help :smile:
 

Answers and Replies

  • #2
335
4
kaitamasaki said:
i understand the basic lim x->0 (sinx/x) = 1 thimg, but if its something more advanced im clueless, such as:

lim 3x^2/(1-cosx) where x->0

how would you solve this one, or something like lim sin(cosx)/cosx where x-> pi/2?

thanks for any help :smile:
Two ways, the first is more general, the second is more useful in this case.
1) Whenever you have a limit that turns into 0/0, use L'Hospital's Rule: Take the derivative of both the numerator and the denominator, then take your limit. So [tex] \lim_{x \rightarrow 0} \frac{3x^2}{1-cosx}= \lim_{x \rightarrow 0} \frac{6x}{sinx} [/tex], which gives 0/0, so do it again: [tex] \lim_{x \rightarrow 0} \frac{6x}{sinx} = \lim_{x \rightarrow 0} \frac{6}{cosx} = 6 [/tex].

2) Use the series expansion for cosine about zero: [tex] cosx = 1 -(1/2!)x^2+(1/4!)x^4-... [/tex]. Take the leading two terms in the cosine expansion and your limit turns into [tex] \lim_{x \rightarrow 0} \frac{3x^2}{1-cosx} = \lim_{x \rightarrow 0} \frac{3x^2}{(1/2)x^2} =2*3=6 [/tex].

-Dan
 
  • #3
VietDao29
Homework Helper
1,423
2
topsquark said:
Two ways, the first is more general, the second is more useful in this case.
1) Whenever you have a limit that turns into 0/0, use L'Hospital's Rule: Take the derivative of both the numerator and the denominator, then take your limit. So [tex] \lim_{x \rightarrow 0} \frac{3x^2}{1-cosx}= \lim_{x \rightarrow 0} \frac{6x}{sinx} [/tex], which gives 0/0, so do it again: [tex] \lim_{x \rightarrow 0} \frac{6x}{sinx} = \lim_{x \rightarrow 0} \frac{6}{cosx} = 6 [/tex].
If you haven't covered L'Hopital rules yet, I'd suggest another way. By the way, L'Hopital's rule is overkill, you should not use it, though...
Since you know that:
[tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex]. You should change your limit a little to have the form above:
[tex]\lim_{x \rightarrow 0} \frac{3x ^ 2}{1 - \cos x}[/tex]
Since you want sin x, and you have (1 - cos x) in the denominator, you may want to multiply both numerator, and denominator by (1 + cos x)
[tex]\lim_{x \rightarrow 0} \frac{3x ^ 2}{1 - \cos x} = \lim_{x \rightarrow 0} \frac{3x ^ 2 (1 + \cos x)}{(1 + \cos x) (1 - \cos x)} = \lim_{x \rightarrow 0} \frac{3x ^ 2 (1 + \cos x)}{\sin ^ 2 x} = 3 \lim_{x \rightarrow 0} \left( \frac{x}{\sin x} \right) ^ 2 (1 + \cos x)[/tex]
Now can you go from here?
kaitamasaki said:
how would you solve this one, or something like lim sin(cosx)/cosx where x-> pi/2?
[tex]\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin (\cos x)}{\cos x}[/tex]
When [tex]x \rightarrow \frac{\pi}{2}[/tex], [tex]\cos x \rightarrow 0[/tex], right? So you'll have the indeterminate form 0 / 0, right?
Now, if you let u = cos x, so as x tends to [tex]\frac{\pi}{2}[/tex], u will tend to 0, right?
So you'll have:
[tex]\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin (\cos x)}{\cos x} = \lim_{u \rightarrow 0} \frac{\sin u}{u} = ?[/tex]
Can you go from here? :)
 

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