Solving Advanced Limits: 3x^2/(1-cosx) & sin(cosx)/cosx

[kaitamasaki]

i understand the basic lim x->0 (sinx/x) = 1 thimg, but if its something more advanced I am 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

 

[topsquark]

i understand the basic lim x->0 (sinx/x) = 1 thimg, but if its something more advanced I am 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

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 \lim_{x \rightarrow 0} \frac{3x^2}{1-cosx}= \lim_{x \rightarrow 0} \frac{6x}{sinx}, which gives 0/0, so do it again: \lim_{x \rightarrow 0} \frac{6x}{sinx} = \lim_{x \rightarrow 0} \frac{6}{cosx} = 6.

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

-Dan

 

[VietDao29]

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 \lim_{x \rightarrow 0} \frac{3x^2}{1-cosx}= \lim_{x \rightarrow 0} \frac{6x}{sinx}, which gives 0/0, so do it again: \lim_{x \rightarrow 0} \frac{6x}{sinx} = \lim_{x \rightarrow 0} \frac{6}{cosx} = 6.

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:
\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1. You should change your limit a little to have the form above:
\lim_{x \rightarrow 0} \frac{3x ^ 2}{1 - \cos x}
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)
\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)
Now can you go from here?

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

\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin (\cos x)}{\cos x}
When x \rightarrow \frac{\pi}{2}, \cos x \rightarrow 0, right? So you'll have the indeterminate form 0 / 0, right?
Now, if you let u = cos x, so as x tends to \frac{\pi}{2}, u will tend to 0, right?
So you'll have:
\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin (\cos x)}{\cos x} = \lim_{u \rightarrow 0} \frac{\sin u}{u} = ?
Can you go from here? :)