Question

How do you solve these limits?

Answer 1

\[\lim_{x \rightarrow \frac{ \pi }{ 2 }} \frac{ 1-\sin ^{3}x }{ 2\cos ^{2}x }\]and\[\lim_{x \rightarrow \frac{ \pi }{ 2 }}\frac{ \sqrt{1-sinx} }{ x-\frac{ \pi }{ 2 } }\]

Answer 2

Do you know about L'Hospital's rule?

Answer 3

I don't remember it but I know what you are talking about.

Answer 4

Both of these limits have both numerator and denominator = 0 at the limit, right, giving us 0/0. L'Hospital says we can find the limit by taking the derivative of both numerator and denominator and taking the limit of that fraction.

Answer 5

ah ok. Thanks

Answer 6

Can you show me how to apply the rule on both limts please?

Answer 7

Stupid buggy website!

The numerator of the fraction is \[1-\sin^3x\]Take the derivative of that with the chain rule:\[u=\sin x\]\[du = \cos x dx\]\[\frac{d}{dx}[1-\sin^3x] = \frac{d}{du}[1-u^3]*\frac{du}{dx} = 0-\cos x*3u^2 = -3\cos x \sin^2x\]

The denominator of the fraction is\[2\cos^2x\]Take the derivative of that with the chain rule:\[u=\cos x\]\[du = -\sin x dx\]\[\frac{d}{dx}[2\cos^x] = -\sin x*2*2u = -4\sin x\cos x\]

Answer 8

Our new limit is \[\lim_{x \rightarrow \pi/2}\frac{-3\cos x\sin^2 x }{ -4 \sin x \cos x}\]You should be able to evaluate that after canceling common factors.

Answer 9

ok, I got that =) And the other limit?

Answer 10

Same deal. Why don't you try it and I'll check your work...

Answer 11

The second limit doesn't have a two-sided limit, as you can see that the value is equal but opposite in sign depending on whether you approach from the + or - side of pi/2...