# Limit of (x^2 - 1)sin(1/(x-1)) as x approaches 1Watch

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#1
Can anybody show me how to proceed with this question?

So far I've noticed that:

However, I'm unsure of how to proceed with the limit of the sine function.
0
5 years ago
#2
(Original post by Khallil)
Can anybody show me how to proceed with this question?

So far I've noticed that:

However, I'm unsure of how to proceed with the limit of the sine function.
We have xsin(1/x) -> 0 as x->0 (because it's bounded by |x| in absolute value). Use x^2 - 1 = (x+1)(x-1).

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#3
(Original post by Hodor)
We have xsin(1/x) -> 0 as x->0 (because it's bounded by |x| in absolute value). Use x^2 - 1 = (x+1)(x-1).

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May I ask how I'm supposed to use the first part? Also I don't understand what's written in your parentheses.
0
5 years ago
#4
(Original post by Khallil)
May I ask how I'm supposed to use the first part? Also I don't understand what's written in your parentheses.
Sure.

For what I put in brackets: we have 0 =< |xsin(1/x)| = |x| |sin(1/x)| =< |x| because |sin(u)| =< 1 for every u. So, since |x| -> 0 as x -> 0, we get xsin(1/x) -> 0 too.

Then, by the above, (x-1)sin(1/(x-1)) -> 0 as x -> 1.

Now: (x^2 - 1)sin(1/(x-1)) = (x+1) (x-1)sin(1/(x-1)). Combining the above with the fact that (x+1) is bounded for x near 1, you should be able to get the answer.

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#5
According to what you've written down (all of with which I agree),

(Original post by Hodor)
Sure.

For what I put in brackets: we have 0 =< |xsin(1/x)| = |x| |sin(1/x)| =< |x| because |sin(u)| =< 1 for every u. So, since |x| -> 0 as x -> 0, we get xsin(1/x) -> 0 too.

Then, by the above, (x-1)sin(1/(x-1)) -> 0 as x -> 1.

Now: (x^2 - 1)sin(1/(x-1)) = (x+1) (x-1)sin(1/(x-1)). Combining the above with the fact that (x+1) is bounded for x near 1, you should be able to get the answer.

Also is this acceptable?
I made the hand wavey argument that

to deduce that

0
5 years ago
#6
(Original post by Khallil)
According to what you've written down (all of with which I agree),

Is this correct?

I made the hand wavey argument that
Looks good to me!

I think you're fine to assume that |sin(u)| =< 1 when doing this sort of thing, but you can prove it in one line using Pythagoras (for example) if you feel it's needed.

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5 years ago
#7
(Original post by Khallil)
According to what you've written down (all of with which I agree),

Also is this acceptable?
I made the hand wavey argument that

to deduce that

Yes, I think your other argument is fine. It's pretty much the same thing. I made things more complicated than necessary.

General principle: if f(x) goes to 0 as x goes to c and g(x) is bounded on an open interval containing c, then f(x)g(x) also goes to 0 as x goes to c.

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