Limit x→1 sin(x-1)/(x²+x-2)?
So my teacher claims the answer is 1/3, but how can that be?
My REAL question is, does the identity {[ Limit x→0 sin(x)/x ] = 1} apply here? Doesn't the identity only work when x approaches 0? In this particular question, x approaches 1. Therefore, the identity is not applicable, thus giving DNE as the solution? Or no?
Update:Second question: Does the identity sin(x)/x = 1 when limit approaches 0 ONLY when it approaches 0? Or is the number interchangeable? If that makes sense..
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lim(x→1) sin(x-1)/(x-1) =1
Rewriting it the way below would help you understand.
Let y = x-1, then x = y +1
lim(x→1) sin(x-1)/(x-1) would become
lim(y+1→1) sin(y) / y
That is, lim(y→0) sin(y) / y, which you know is 1.
Now back to your question.
(x²+x-2) = (x-1)(x+2)
lim(x→1) sin(x-1)/(x²+x-2)
=lim(x→1) sin(x-1) / ((x-1)(x+2))
=lim(x→1) sin(x-1)/(x-1) * lim(x→1) 1/(x+2), you can do this when both limits exist!
=1*1/(1+2)
=1/3
Hope this clears you up a little.
L=lim{x->1} [sin(x-1)/(x^2+x-2)], assuming it exists
Since this is of form 0/0, we get by L'hospitals Rule
L=lim{x->1} [d/dx(sin(x-1)/(d/dx(x^2+x-2))]
L=lim{x->1} [cos(x-1)/(2x+1)]
If we now plug in x=1
L=cos(1-1)/(2(1)+1)=cos(0)/3=1/3