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Differentiability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiability5 Marks
Question
If $\text{f(x)}=\sqrt{\text{x}^2+9},$ Write the value of $\lim\limits_{\text{x}\rightarrow4}\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}.$
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Answer

Given: $\text{f(x)}=\text{x}^2+9$
Now,
$\text{f}(4)=\sqrt{16+9}$
$=\sqrt{25}$
$=5$
So,
$\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}=\frac{\sqrt{\text{x}^2+9-5}}{\text{x}-4}$
On rationalising the numeratore, we get
$\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}$
$=\frac{\sqrt{\text{x}^2+9-5}}{\text{x}-4}\times\frac{\sqrt{\text{x}^2+9}+5}{\sqrt{\text{x}^2+9}+5}$
$=\frac{\text{x}^2+9-25}{(\text{x}-4)(\sqrt{\text{x}^2+9+5})}$
$=\frac{\text{x}^2-16}{(\text{x}-4)(\sqrt{\text{x}^2+9+5})}$
$=\frac{\text{x}+4}{\sqrt{\text{x}^2+9+5}}$
Taking limit x → 4, we have
$\lim\limits_{\text{x}\rightarrow4}\frac{\text{f(x)}-\text{f}(4)}{\text{x}-4}=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}+4}{\sqrt{\text{x}^2+9+5}}$
$=\frac{8}{10}$
$=\frac{4}{5}$

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