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प्रश्न
Discuss the continuity and differentiability of f (x) = e|x| .
उत्तर
Discuss the continuity and differentiability of f (x) = e|x| .
Given:
⇒f`(x) = {(e^x, ,xge0),(e^(-x), ,x<0):}`
Continuity:
(LHL at x = 0)
\[\lim_{x \to 0^-} f(x) \]
\[ = \lim_{h \to 0} f(0 - h) \]
\[ = \lim_{h \to 0} e^{- (0 - h)} \]
\[ = \lim_{h \to 0} e^h \]
\[ = 1\]
(RHL at x = 0)
\[\lim_{x \to 0^+} f(x) \]
\[ = \lim_{h \to 0} f(0 + h) \]
\[ = \lim_{h \to 0} e^{(0 + h)} \]
\[ = 1\]
Thus,
Hence,function is continuous at x = 0 .
Differentiability at x = 0.
(LHD at x = 0)
\[\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}\]
\[ = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0}\]
\[ = \lim_{h \to 0} \frac{e^{- (0 - h)} - 1}{- h}\]
\[ = \lim_{h \to 0} \frac{e^h - 1}{- h} \]
\[ = - 1 \left[ \because \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \right]\]
(RHD at x = 0)
\[\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]
\[ = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{0 + h - 0}\]
\[ = \lim_{h \to 0} \frac{e^{(0 + h)} - 1}{h}\]
\[ = \lim_{h \to 0} \frac{e^h - 1}{h} \]
\[ = 1 \left[ \because \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \right]\]
LHD at (x = 0)
Hence the function is not differentiable at x = 0.
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