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

Khallil
Badges: 13
Rep:
?
#1
Report Thread starter 5 years ago
#1
Can anybody show me how to proceed with this question?

\text{Evaluate} \ \displaystyle \lim_{x \to 1} \left( \left(x^2 - 1 \right) \sin \left( \frac{1}{x-1} \right) \right)

So far I've noticed that:

\displaystyle \lim_{x \to 1} \left(x^2 - 1 \right) = 0

However, I'm unsure of how to proceed with the limit of the sine function.
0
reply
Hodor
Badges: 3
Rep:
?
#2
Report 5 years ago
#2
(Original post by Khallil)
Can anybody show me how to proceed with this question?

\text{Evaluate} \ \displaystyle \lim_{x \to 1} \left( \left(x^2 - 1 \right) \sin \left( \frac{1}{x-1} \right) \right)

So far I've noticed that:

\displaystyle \lim_{x \to 1} \left(x^2 - 1 \right) = 0

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).


Posted from TSR Mobile
1
reply
Khallil
Badges: 13
Rep:
?
#3
Report Thread starter 5 years ago
#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).


Posted from TSR Mobile
May I ask how I'm supposed to use the first part? Also I don't understand what's written in your parentheses.
0
reply
Hodor
Badges: 3
Rep:
?
#4
Report 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.


Posted from TSR Mobile
0
reply
Khallil
Badges: 13
Rep:
?
#5
Report Thread starter 5 years ago
#5
According to what you've written down (all of with which I agree),

\displaystyle \begin{aligned} \lim_{x \to 1} \ \left[ (x^2 - 1) \sin \left( \frac{1}{x-1} \right) \right] & = \lim_{x \to 1} \ \left[ (x+1) (x-1) \sin \left( \frac{1}{x-1} \right) \right] \\ & = \lim_{x \to 1} \ (x+1) \cdot \lim_{x \to 1} \ \left[ (x-1)\sin \left( \frac{1}{x-1} \right) \right] \\ & = 2 \times 0 = 0 \end{aligned}

(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

\forall x: \ -1 \leq \sin(x) \leq 1

to deduce that

\displaystyle \lim_{x \to 1} \left( x^2 - 1 \right) \sin \left( \frac{1}{x-1} \right) = 0 \times \left( \text{any value in} \ \left[-1, 1 \right] \right) = 0
0
reply
Hodor
Badges: 3
Rep:
?
#6
Report 5 years ago
#6
(Original post by Khallil)
According to what you've written down (all of with which I agree),

\displaystyle \begin{aligned} \lim_{x \to 1} \ \left[ (x^2 - 1) \sin \left( \frac{1}{x-1} \right) \right] & = \lim_{x \to 1} \ \left[ (x+1) (x-1) \sin \left( \frac{1}{x-1} \right) \right] \\ & = \lim_{x \to 1} \ (x+1) \cdot \lim_{x \to 1} \ \left[ (x-1)\sin \left( \frac{1}{x-1} \right) \right] \\ & = 2 \times 0 = 0 \end{aligned}

Is this correct?

I made the hand wavey argument that \forall x: \ -1 \leq \sin(x) \leq 1
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.


Posted from TSR Mobile
0
reply
Hodor
Badges: 3
Rep:
?
#7
Report 5 years ago
#7
(Original post by Khallil)
According to what you've written down (all of with which I agree),

\displaystyle \begin{aligned} \lim_{x \to 1} \ \left[ (x^2 - 1) \sin \left( \frac{1}{x-1} \right) \right] & = \lim_{x \to 1} \ \left[ (x+1) (x-1) \sin \left( \frac{1}{x-1} \right) \right] \\ & = \lim_{x \to 1} \ (x+1) \cdot \lim_{x \to 1} \ \left[ (x-1)\sin \left( \frac{1}{x-1} \right) \right] \\ & = 2 \times 0 = 0 \end{aligned}




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

\forall x: \ -1 \leq \sin(x) \leq 1

to deduce that

\displaystyle \lim_{x \to 1} \left( x^2 - 1 \right) \sin \left( \frac{1}{x-1} \right) = 0 \times \left( \text{any value in} \ \left[-1, 1 \right] \right) = 0
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.

Posted from TSR Mobile
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • London Metropolitan University
    Postgraduate Mini Open Evening - Holloway Campus Undergraduate
    Tue, 21 May '19
  • Brunel University London
    Postgraduate Open Evening Postgraduate
    Wed, 22 May '19
  • University of Roehampton
    Postgraduate Open Evening Postgraduate
    Wed, 22 May '19

How did your AQA A-level Economics Paper 1 go?

Loved the paper - Feeling positive (53)
14.29%
The paper was reasonable (187)
50.4%
Not feeling great about that exam... (90)
24.26%
It was TERRIBLE (41)
11.05%

Watched Threads

View All