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Lim X → 5 X − 5 √ 6 X − 5 − √ 4 X + 5

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प्रश्न

\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\] 

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उत्तर

\[\lim_{x \to 5} \left[ \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}} \right]\] It is of the form \[\frac{0}{0}\] 

Rationalising the denominator: 

\[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{\left( \sqrt{6x - 5} - \sqrt{4x + 5} \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)} \right]\] 

=  \[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{\left( 6x - 5 \right) - \left( 4x + 5 \right)} \right]\] 

=  \[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right) \left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{2x - 10} \right]\] 

=\[\lim_{x \to 5} \left[ \frac{\left( x - 5 \right)\left( \sqrt{6x - 5} + \sqrt{4x + 5} \right)}{2\left( x - 5 \right)} \right]\] 

=  \[\frac{\sqrt{6 \times 5 - 5} + \sqrt{4 \times 5 + 5}}{2}\] 

= \[\frac{5 + 5}{2}\] 

= 5

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अध्याय 29: Limits - Exercise 29.4 [पृष्ठ २८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.4 | Q 17 | पृष्ठ २८

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