Define f: R rightarrow R by f(x) = begin{case | vedclass

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(C) Given $f(x) = (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}$ for $x 
eq a$ and $f(a) = 0$. To check continuity at $x=a$,we evaluate the l

Vedclass · Accepted Answer

The function $f(x)$ is continuous at $x=a$

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(C) Given $f(x) = (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}$ for $x 
eq a$ and $f(a) = 0$. To check continuity at $x=a$,we evaluate the limits: Left-hand limit: $\lim_{x ightarrow a^-} f(x) = \lim_{h ightarrow 0} (-h) \frac{e^{-1/h}-1}{e^{-1/h}+1} = 0 	imes \frac{0-1}{0+1} = 0$. Right-hand limit: $\lim_{x ightarrow a^+} f(x) = \lim_{h ightarrow 0} (h) \frac{e^{1/h}-1}{e^{1/h}+1} = \lim_{h ightarrow 0} h \frac{1-e^{-1/h}}{1+e^{-1/h}} = 0 	imes \frac{1-0}{1+0} = 0$. Since $\lim_{x ightarrow a^-} f(x) = \lim_{x ightarrow a^+} f(x) = f(a) = 0$,the function $f(x)$ is continuous at $x=a$.

Define $f: R \rightarrow R$ by $f(x) = \begin{cases} (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}, & x \neq a \\ 0, & x=a \end{cases}$. Then which one of the following is true?

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