If lim _{x rightarrow 0}left(frac{1+cx}{1-cx} | vedclass

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(C) We know that $\lim _{x \rightarrow 0} (1+ax)^{1/x} = e^a$.Using this, $\lim _{x \rightarrow 0} \left(\frac{1+cx}{1-cx}\right)^{1/x} = \frac{\lim _

Vedclass · Accepted Answer

$16$

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(C) We know that $\lim _{x \rightarrow 0} (1+ax)^{1/x} = e^a$.Using this, $\lim _{x \rightarrow 0} \left(\frac{1+cx}{1-cx}\right)^{1/x} = \frac{\lim _{x \rightarrow 0} (1+cx)^{1/x}}{\lim _{x \rightarrow 0} (1-cx)^{1/x}} = \frac{e^c}{e^{-c}} = e^{2c}$.Given $e^{2c} = 4$.Now, we need to find $\lim _{x \rightarrow 0} \left(\frac{1+2cx}{1-2cx}\right)^{1/x} = \frac{\lim _{x \rightarrow 0} (1+2cx)^{1/x}}{\lim _{x \rightarrow 0} (1-2cx)^{1/x}} = \frac{e^{2c}}{e^{-2c}} = e^{4c}$.Since $e^{2c} = 4$, then $e^{4c} = (e^{2c})^2 = 4^2 = 16$.

If $\lim _{x \rightarrow 0}\left(\frac{1+cx}{1-cx}\right)^{1/x}=4$, then $\lim _{x \rightarrow 0}\left(\frac{1+2cx}{1-2cx}\right)^{1/x}$ is

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