If f(x) = begin{cases} (1 + 2x)^{1/x}, & x ne | vedclass

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(B) To check the continuity at $x = 0$,we calculate the limit of $f(x)$ as $x 	o 0$.Given $f(x) = (1 + 2x)^{1/x}$ for $x 
e 0$.We know the standard

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$\lim_{x 	o 0^-} f(x) = e^2$

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(B) To check the continuity at $x = 0$,we calculate the limit of $f(x)$ as $x 	o 0$.Given $f(x) = (1 + 2x)^{1/x}$ for $x 
e 0$.We know the standard limit $\lim_{u 	o 0} (1 + u)^{1/u} = e$.Here,let $u = 2x$. As $x 	o 0$,$u 	o 0$.Then $f(x) = (1 + 2x)^{1/x} = [(1 + 2x)^{1/(2x)}]^2$.Taking the limit as $x 	o 0$:$\lim_{x 	o 0} f(x) = \lim_{u 	o 0} [(1 + u)^{1/u}]^2 = e^2$.Since $\lim_{x 	o 0} f(x) = e^2$ and $f(0) = e^2$,the function is continuous at $x = 0$.Thus,$\lim_{x 	o 0^+} f(x) = e^2$ and $\lim_{x 	o 0^-} f(x) = e^2$.Comparing with the options,option $(b)$ is correct.

If $f(x) = \begin{cases} (1 + 2x)^{1/x}, & x \ne 0 \\ e^2, & x = 0 \end{cases}$,then:

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