Question

45-60: Find the limit.

60. \(\mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^2} + x - 2}}\)

Short answer

The required value is \(\mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^2} + x - 2}} = \frac{1}{3}\).

Step-by-step solution

Special Trigonometric limits

The special trigonometric for sine function is given below:

\(\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1\)

Limit evaluation

The required limit is\(\mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^2} + x - 2}}\).

Manipulate and solve the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^2} + x - 2}} &= \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^2} + x - 2}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{\left( {x - 1} \right)}} \cdot \frac{1}{{\left( {x + 2} \right)}}\\ &= \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{\left( {x - 1} \right)}} \cdot \mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {x + 2} \right)}}\end{aligned}\)

Solve the above limit further as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^2} + x - 2}} &= \mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{\left( {x - 1} \right)}} \cdot \mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {x + 2} \right)}}\\ &= \mathop {\lim }\limits_{x - 1 \to 0} \frac{{\sin \left( {x - 1} \right)}}{{\left( {x - 1} \right)}} \cdot \mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {x + 2} \right)}}\\ &= 1 \cdot \frac{1}{{\left( {1 + 2} \right)}}\\ &= \frac{1}{3}\end{aligned}\)

Hence, \(\mathop {\lim }\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^2} + x - 2}} = \frac{1}{3}\).