मराठी

lim x → 0 2 x √ a + x − √ a − x

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प्रश्न

\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{2x}{\sqrt{a + x} - \sqrt{a - x}} \right]\] 

When x = 0, then the expression \[\frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\] becomes \[\frac{0}{0}\]  Rationalising the denominator: 

\[\lim_{x \to 0} \left[ \frac{2x}{\left( \sqrt{a + x} - \sqrt{a - x} \right)} \times \frac{\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a + x} + \sqrt{a - x} \right)} \right]\] 

\[\lim_{x \to 0} \left[ \frac{\left( 2x \right)\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( a + x \right) - \left( a - x \right)} \right]\] 

\[\lim_{x \to 0} \left[ \frac{2x\left( \sqrt{a + x} + \sqrt{a - x} \right)}{2x} \right]\] 

=  \[\sqrt{a} + \sqrt{a}\]

\[2\sqrt{a}\] 

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पाठ 29: Limits - Exercise 29.4 [पृष्ठ २८]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.4 | Q 2 | पृष्ठ २८

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