I need to see the step by step work for this. I looked it up on wolfram alpha, and it says the limit is 2, but we haven't learned L'Hospital's rule yet, so I can't use the step by step on wolfram alpha. Could you help please?
kb
Favorite Answer
Use lim(t→0) sin(t)/t = 1 as follows:
lim(x→1) sin(x^2 - 1)/(x - 1)
= lim(x→1) (x + 1) * sin(x^2 - 1)/[(x + 1) * (x - 1)]
= lim(x→1) (x + 1) * sin(x^2 - 1)/(x^2 - 1)
= lim(x→1) (x + 1) * lim(x→1) sin(x^2 - 1)/(x^2 - 1)
= lim(x→1) (x + 1) * lim(t→0) sin(t)/t, letting t = x^2 - 1
= 2 * 1
= 2.
I hope this helps!