Limx - -1e -x1-xx is equal to

Lim_x → ∞( 1e x1+x)^x = lim_x → ∞(1e 11x+1) = (1e 1 nbsp; )^∞ = (some negative nbsp; value)^∞ nbsp; which is not defined as base is negative. nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Indeterminate Forms

limx → ∞1e -x...

Question

$lim x \to \infty {(\frac{1}{e} - \frac{x}{1 + x})}^{x}$ is equal to

\frac{e}{1 - e}

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0

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\frac{e}{e^{1 - e}}

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Does not exist

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Solution

The correct option is D Does not exist
$lim x \to \infty {(\frac{1}{e} - \frac{x}{1 + x})}^{x} = lim x \to \infty ⎛ ⎜ ⎜ ⎜ ⎝ \frac{1}{e} - \frac{1}{\frac{1}{x} + 1} ⎞ ⎟ ⎟ ⎟ ⎠ = {(\frac{1}{e} - 1)}^{\infty} = (some negative value)^{\infty}$
which is not defined as base is negative.

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limx → ∞(1e−x1+x)x is equal to

The correct option is D Does not exist limx → ∞(1e−x1+x)x=limx → ∞⎛⎜ ⎜ ⎜⎝1e−11x+1⎞⎟ ⎟ ⎟⎠=(1e−1)∞=(some negative value)∞ which is not defined as base is negative.

$lim x \to \infty {(\frac{1}{e} - \frac{x}{1 + x})}^{x}$ is equal to