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Find the limit of lim(x->1)[sin(x-1)/(x^2 +x-2)]. I tried the question, but is the answer 1/3?
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\[\lim_{x\to1}\frac{sin(x-1)}{x^2 + x-2}\]
I will solve the denominator first \[\begin{array}r x^2 +x -1 -1 = x^2 - 1 +x-1 = (x-1)(x+1) + (x-1) \\ \implies (x -1)(x+2)\end{array}\]
\[\lim_{x \to1} \frac{sin(x-1)}{x-1} \times \frac{1}{x+2}\] \[ 1 \times \frac{1}{3}\]
How does lim(x->1)[sin(x-1)/(x-1)] become 1? I don't get it.. our prof told us that lim(x->0)[sinx/x]=1 , so i assumed that the answer can be 1 only if the x approaches 0..
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