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Lim X → 7 4 − √ 9 + X 1 − √ 8 − X - Mathematics

\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\] 

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Solution

\[\lim_{x \to 7} \left[ \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}} \right]\] It is of the form \[\frac{0}{0}\] 

Rationalising the numerator and the denominator: 

\[\lim_{x \to 7} \left[ \frac{\left( 4 - \sqrt{9 + x} \right)}{1} \times \frac{\left( 4 + \sqrt{9 + x} \right)}{\left( 4 + \sqrt{9 + x} \right)} \times \frac{1}{\left( 1 - \sqrt{8 - x} \right)} \times \frac{\left( 1 + \sqrt{8 - x} \right)}{\left( 1 + \sqrt{8 - x} \right)} \right]\] 

=  \[\lim_{x \to 7} \left[ \frac{16 - \left( 9 + x \right)}{\left( 4 + \sqrt{9 + x} \right)} \times \frac{\left( 1 + \sqrt{8 - x} \right)}{1 - \left( 8 - x \right)} \right]\] 

=  \[\lim_{x \to 7} \left[ \frac{- 1\left( - 7 + x \right)\left( 1 + \sqrt{8 - x} \right)}{\left( 4 + \sqrt{9 + x} \right)\left( - 7 + x \right)} \right]\] 

=  \[\lim_{x \to 7} \left[ \frac{- \left( 1 + \sqrt{8 - x} \right)}{4 + \sqrt{9 + x}} \right]\] 

= \[- \left( \frac{1 + \sqrt{8 - 7}}{4 + \sqrt{9 + 7}} \right)\]

=  \[\frac{- 2}{4 + 4}\] 

=  \[\frac{- 1}{4}\] 

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.4 | Q 15 | Page 28
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