The right hand and left hand limit of the fun | vedclass

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(B) Given $f(x) = \begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & x 
eq 0 \ 0, & x=0 \end{cases}$ Right hand limit $(RHL)$: $\lim_{x ightarrow 0^

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$1$ and $-1$

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(B) Given $f(x) = \begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & x 
eq 0 \ 0, & x=0 \end{cases}$ Right hand limit $(RHL)$: $\lim_{x ightarrow 0^{+}} f(x) = \lim_{h ightarrow 0} f(0+h) = \lim_{h ightarrow 0} \frac{e^{1/h}-1}{e^{1/h}+1}$ Dividing numerator and denominator by $e^{1/h}$: $\lim_{h ightarrow 0} \frac{1 - e^{-1/h}}{1 + e^{-1/h}} = \frac{1 - 0}{1 + 0} = 1$ Left hand limit $(LHL)$: $\lim_{x ightarrow 0^{-}} f(x) = \lim_{h ightarrow 0} f(0-h) = \lim_{h ightarrow 0} \frac{e^{-1/h}-1}{e^{-1/h}+1}$ As $h ightarrow 0^{+}$,$e^{-1/h} ightarrow 0$: $\frac{0 - 1}{0 + 1} = -1$ Thus,the $RHL$ is $1$ and the $LHL$ is $-1$.

The right hand and left hand limit of the function $f(x)$ are respectively:
$f(x)=\begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

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The right hand and left hand limit of the function $f(x)$ are respectively: $f(x)=\begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$