Question

\[\lim_{x \rightarrow 1} \frac{\sin(x^2 - 1)}{x - 1}\]
I factored the top into \[\lim_{x \rightarrow 1} \frac{\sin((x - 1)(x + 1))}{x - 1}\]
but now what

Answer 1

Both the top and bottom tend to 0 as x tends to 1, so you can use L'Hopital's rule. Have you used this before?

Answer 2

No, we're not allowed to use l'Hopital

Answer 3

differnetiate num and den seprately and put the limits

Answer 4

why is that so...

Answer 5

What have you been doing in class? You could do it by series or some other way.

Answer 6

It's an introductory Calculus 1 course. If we use any method not taught in class, we get no credit (someone used differentiation laws on a limit problem and got a 0 because we didn't learn the differentiation rules yet at the time)

Answer 7

I know that sinx/x and x/sinx = 1, so I'm trying to turn it into that familiar form, but I can't seem to do it because the x^2 - 1 is within the sin and in the bottom I only have x - 1

Answer 8

Oh wait, I think I just got it

Answer 9

multiply numerator and denom by x+1

Answer 10

Multiply both numerator and denominator by (x + 1) and you get 1 + 1

Answer 11

yup, u get 2.

Answer 12

That's one way, or if you've done Taylor expansions that way would work.

Answer 13

Isn't Taylor expansions like in Calc 3 or something

Answer 14

Can I ask a question?

Answer 15

I don't know what your syllabus is teaching, I did it in Calculus101.

Answer 16

Go ahead @sauravshakya

Answer 17

How?
sinx/x=1

Answer 18

there are many ways to prove it.....series expansion, approximation, geometric way.....

Answer 19

I think this only works when x---->0

Answer 20

Yeh it's a nice property, and yes it is only as x->0.

Answer 21

yes, absolutely, we are talking about limits only here, isn't it ?

Answer 22

But if you have x->1 and (sin(x-1))(x-1) then you can label y=x-1 so you have \(\frac{siny}{y}\) for y->0.

Answer 23

I mean i got it

Answer 24

Thanx guys