limit sin(x-1)\x^2-1?

Question

it's professional math.

Answer 1

limit sin(x-1)\x^2-1 as x->1

Use L'Hospital theorem

limit f(x)/g(x) = limit f'(x)/g'(x)

Here

limit sin(x-1)\(x^2-1) =
limit cos(x-1)/(2x) = 1/2

Answer 2

lim [ sin(x - 1) / (x^2 - 1) ]
x -> 1

That's what I'm assuming you mean (because that's the only way this question's answer is nontrivial.

To solve this, note that if we plug in x = 1, we will obtain the form
[0/0]. Therefore, one option is to use L'Hospital's rule. This involves taking the derivative of the top and bottom, and finding this new limit. Doing so, we get

lim [ cos(x - 1) / 2x ]
x -> 1

Plugging x = 1 into this limit, we get

cos(1 - 1) / 2(1) = cos(0) / 2 = 1/2

Answer 3

A=sin(x-1)/(x^2-1)

(x^2-1)=(x-1)(x+1)

=>A=sin(x-1)/{(x-1)(x+1)}={sin(x-1)/(x-1)}{1/(x+1)}

put z=x-1=> x=z+1

=> => A={sinz/z}/(z+2)

i think that you are looking for the lim -->0


lim(sinz/z)=1

limA=1/(0+2)=>
=> limA=1/2


ok?

Answer 4

x^2 -1 = (x+1)*(x-1)

When x→1
Lt sin (x-1) / ( x^2 -1)

= Lt sin (x-1) / (x+1)*(x-1)

=Lt 1/ (x+1)* Lt sin (x-1) / (x-1)

=Lt 1/ (x+1) * 1 = ½.

Answer 5

As x approaches what?

Answer 6

uh...