Question

Limit help!

limit of [((sin^2)x)/x] as x approaches 0

Answer 1

\[\lim_{x\to0}\frac{\sin^2x}{x}=\lim_{x\to0}\sin x\cdot\lim_{x\to0}\frac{\sin x}{x}\]

Answer 2

but when you plug zero into the sin, how do you know which sin to use? there's two sins with zeros

Answer 3

Not sure I understand your question.
\(\sin^2x=\sin x\sin x\), so you have
\[\lim_{x\to0}\frac{\sin x\sin x}{x}\]
Then, using a property of limits, you have
\[\lim_{x\to0}\sin x\cdot\lim_{x\to0}\frac{\sin x}{x}\]
You have two limits here. The first is 0, since \(\sin0=0\). The second is 1, since \(\lim_{x\to0}\dfrac{\sin x}{x}=1\), a known limit.

So all-in-all, your answer is 0, since 0∙1=0.

Answer 4

Would you need the unit circle to find the sin of zero or is that just something I should know?

Answer 5

Yeah, you should be familiar with \(\sin0=0\). Do you know about the second limit though? I would think that's the more important aspect of the problem.

Answer 6

..a second limit?

Answer 7

The sinx/x one, I mean. Have you learned that
\[\lim_{x\to0}\frac{\sin x}{x}=1~~?\]

Answer 8

Yes, and I also learned that (1-cosx)/x = 1

Answer 9

whoops! that should be = 0

Answer 10

Right, well this problem is just making you apply that knowledge.

Answer 11

Could you help me with another limit problem?

Answer 12

Yeah

Answer 13

what is the limit of (cos x/cot x) as x approaches pi/2 ?

Answer 14

\[\frac{\cos x}{\cot x}=\frac{\cos x}{\frac{\cos x}{\sin x}}=\cos x\frac{\sin x}{\cos x}=\sin x\]
So what's the limit
\[\lim_{x\to\pi/2}\sin x~~?\]

Answer 15

wait, when you changed the cot x in the denominator, what did you change it to? I can't see it xD

Answer 16

\[\cot x=\frac{\cos x}{\sin x}\]

Answer 17

the sin of pi/2 is 1

Answer 18

Correct!

Answer 19

I have one last problem if you don't mind?