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Lim X → 2 X 2 − X − 2 X 2 − 2 X + Sin ( X − 2 )

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प्रश्न

\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\] 

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उत्तर

\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[ = \lim_{x \to 2} \frac{\left( x - 2 \right) \left( x + 1 \right)}{x\left( x - 2 \right) + \sin \left( x - 2 \right)}\]
\[\text{ Let } y = x - 2\]
\[ x \to 2\]
\[ \therefore y \to 0\]
\[ = \lim_{y \to 0} \frac{y\left( y + 3 \right)}{\left( y + 2 \right)y + \sin y}\]
\[\text{ Dividing the numerator and the denominator by } y:\]
\[ = \lim_{y \to 0} \frac{\left( y + 3 \right)}{\left( y + 2 \right) + \frac{\sin y}{y}}\]
\[ = \frac{3}{2 + 1}\]
\[ = \frac{3}{3} = 1\]

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 28 | पृष्ठ ६२

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