Question

b) (lim_{x \to 1} \frac{sin(x - 1)}{x^2 + x - 2})

Explanation:

Step1: Factor the denominator

First, we factor the quadratic expression in the denominator. For \(x^2 + x - 2\), we find two numbers that multiply to \(-2\) and add to \(1\). Those numbers are \(2\) and \(-1\), so we can factor it as:
\(x^2 + x - 2=(x + 2)(x - 1)\)
So the limit becomes:
\(\lim_{x
ightarrow1}\frac{\sin(x - 1)}{(x + 2)(x - 1)}\)

Step2: Rewrite the expression

We can rewrite the expression as a product of two fractions:
\(\lim_{x
ightarrow1}\frac{\sin(x - 1)}{x - 1}\cdot\frac{1}{x + 2}\)

Step3: Use the standard limit

We know that the standard limit \(\lim_{u
ightarrow0}\frac{\sin(u)}{u}=1\). Let \(u=x - 1\). As \(x
ightarrow1\), \(u
ightarrow0\). So \(\lim_{x
ightarrow1}\frac{\sin(x - 1)}{x - 1}=\lim_{u
ightarrow0}\frac{\sin(u)}{u}=1\)

Step4: Evaluate the remaining limit

Now we evaluate \(\lim_{x
ightarrow1}\frac{1}{x + 2}\). Substituting \(x = 1\) into \(\frac{1}{x + 2}\), we get \(\frac{1}{1+2}=\frac{1}{3}\)

Step5: Multiply the two limits

Since the limit of a product is the product of the limits (when both limits exist), we multiply the results from Step 3 and Step 4:
\(1\times\frac{1}{3}=\frac{1}{3}\)

Answer:

\(\frac{1}{3}\)