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प्रश्न
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 1 + x \right)^6 - 1}{x} \times \frac{x}{\left( 1 + x \right)^2 - 1} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 1 + x \right)^6 - 1^6}{\left( 1 + x \right) - 1} \times \frac{\left( 1 + x \right) - 1}{\left( 1 + x \right)^2 - 1} \right]\]
Let y = 1 + x
When x → 0, then 1 + x → 1.
\[\Rightarrow\]y → 1
\[\lim_{y \to 1} \left[ \left( \frac{y^6 - 1^6}{y - 1} \right) \times \frac{\left( y - 1 \right)}{y^2 - 1^2} \right]\]
\[ = \frac{6 \times \left( 1 \right)^{6 - 1}}{2 \times \left( 1 \right)^{2 - 1}}\]
\[ = \frac{6}{2}\]
\[ = 3\]
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