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

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

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

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

\infty

Solution steps

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

Simplify

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

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

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

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

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

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

= 1 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

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