What is the limit as x approaches 0+of e^{-1/x}ln(x) ?

The limit as x approaches 0+of e^{-1/x}ln(x) is 0

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limit as x approaches 0+of e^{-1/x}ln(x)

x \to 0 + lim (e^{- \frac{1}{x}} ln (x))

0

Solution steps

x \to 0 + lim (e^{- \frac{1}{x}} ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{e ^{- \frac{1}{x}}}})

Apply L'Hopital's Rule

= x \to 0 + lim \frac{\frac{1}{x}}{- \frac{e ^{\frac{1}{x}}}{x ^{2}}}

Simplify

\frac{\frac{1}{x}}{- \frac{e ^{\frac{1}{x}}}{x ^{2}}} : - \frac{x}{e ^{\frac{1}{x}}}

= x \to 0 + lim (- \frac{x}{e ^{\frac{1}{x}}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= - x \to 0 + lim (\frac{x}{e ^{\frac{1}{x}}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= - \frac{lim _{x \to 0 +} ( x )}{lim _{x \to 0 +} ( e ^{\frac{1}{x}} )}

x \to 0 + lim (x) = 0

x \to 0 + lim (e^{\frac{1}{x}}) = \infty

= - \frac{0}{\infty}

Apply Infinity Property:

\frac{0}{\infty} = 0

= 0

What is the limit as x approaches 0+of e^{-1/x}ln(x) ?
The limit as x approaches 0+of e^{-1/x}ln(x) is 0