What is the limit as x approaches 0+of ln(x)+1/(x^2) ?

The limit as x approaches 0+of ln(x)+1/(x^2) is infinity

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limit as x approaches 0+of ln(x)+1/(x^2)

x \to 0 + lim (ln (x) + \frac{1}{x ^{2}})

\infty

Solution steps

x \to 0 + lim (ln (x) + \frac{1}{x ^{2}})

Simplify

ln (x) + \frac{1}{x ^{2}} : \frac{x ^{2} ln ( x ) + 1}{x ^{2}}

= x \to 0 + lim (\frac{x ^{2} ln ( x ) + 1}{x ^{2}})

\frac{x ^{2} ln ( x ) + 1}{x ^{2}} = (x^{2} ln (x) + 1) \frac{1}{x ^{2}}

= x \to 0 + lim ((x^{2} ln (x) + 1) \frac{1}{x ^{2}})

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

With the exception of indeterminate form

= x \to 0 + lim (x^{2} ln (x) + 1) \cdot x \to 0 + lim (\frac{1}{x ^{2}})

x \to 0 + lim (x^{2} ln (x) + 1) = 1

x \to 0 + lim (\frac{1}{x ^{2}}) = \infty

= 1 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

\int sec (14 w) tan (14 w) d w

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n = 0 \sum \infty \frac{1}{4 n ^{2} - 1}

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What is the limit as x approaches 0+of ln(x)+1/(x^2) ?
The limit as x approaches 0+of ln(x)+1/(x^2) is infinity