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If F ( X ) = √ X 2 + 9 , Write the Value of Lim X → 4 F ( X ) − F ( 4 ) X − 4 . - Mathematics

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प्रश्न

If  \[f \left( x \right) = \sqrt{x^2 + 9}\] , write the value of

\[\lim_{x \to 4} \frac{f\left( x \right) - f\left( 4 \right)}{x - 4} .\]
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उत्तर

Given:   

\[f(x) = \sqrt{x^2 + 9}\]

Now,

\[f(4) = \sqrt{16 + 9} \]
\[ = \sqrt{25} \]
\[ = 5\]

So, 

\[\frac{f(x) - f(4)}{x - 4} = \frac{\sqrt{x^2 + 9} - 5}{x - 4}\]

On rationalising the numerator, we get

\[\frac{f(x) - f(4)}{x - 4} = \frac{\sqrt{x^2 + 9} - 5}{x - 4} \times \frac{\sqrt{x^2 + 9} + 5}{\sqrt{x^2 + 9} + 5} \]
\[ = \frac{x^2 + 9 - 25}{(x - 4) \left( \sqrt{x^2 + 9} + 5 \right)} \]
\[ = \frac{x^2 - 16}{(x - 4) \left( \sqrt{x^2 + 9} + 5 \right)}\]
\[ = \frac{(x + 4)}{\sqrt{x^2 + 9} + 5}\]

Taking limit 

\[x \to 4\], we have 
\[\lim_{x \to 4} \frac{f(x) - f(4)}{x - 4} = \lim_{x \to 4} \frac{(x + 4)}{\sqrt{x^2 + 9} + 5} \]
\[ = \frac{8}{10} \]
\[ = \frac{4}{5}\]
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अध्याय 10: Differentiability - Exercise 10.3 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 10 Differentiability
Exercise 10.3 | Q 13 | पृष्ठ १७

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