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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$ exists finitely, write the value of $\lim\limits_{\text{x}\rightarrow{\text{c}}}\text{f(x)}.$
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Answer

LHL = f(1) = RHL
So, f(x) is continuous at x = 1
Now,
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}-(1)}{(1-\text{h})-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-1}{-\text{h}}$
= Not defined
(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}-(1)}{(1+\text{h})-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2(1+\text{h})-1-1}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{h}}$
$=1$
(LHL at x = 1) $\neq$ (RHL at x = 1)
$\therefore$ f(x) is continuous but not differentiable at x = 0 and 1.
$\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$ exists finitely
So,
$\text{f}'(\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$
$\text{f}'(\text{c})\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x}-\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$\text{f}'(\text{c})(\text{c}-\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$0=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$\lim\limits_{\text{x}\rightarrow{\text{c}}}\text{f(x})=\text{f(c)}$

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