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प्रश्न
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
उत्तर
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{x^2 + x + \sin \left( x + 1 \right)}\]
\[ = \lim_{x \to - 1} \frac{\left( x - 2 \right) \left( x + 1 \right)}{x\left( x + 1 \right) + \sin \left( x + 1 \right)}\]
\[Let y = x + 1\]
\[If x \to - 1, then y \to 0 . \]
\[ = \lim_{y \to 0} \frac{\left( y - 3 \right)y}{\left( y - 1 \right)y + \sin y}\]
\[\text{ Dividing the numerator and the denominator by } y:\]
\[ = \lim_{y \to 0} \frac{\left( y - 3 \right)}{\left( y - 1 \right) + \frac{\sin y}{y}}\]
\[ = \frac{0 - 3}{\left( 0 - 1 \right) + 1}\]
\[ \Rightarrow \frac{- 3}{0} = \infty\]
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