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

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

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

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

\infty

Solution steps

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

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

With the exception of indeterminate form

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

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

x \to 0 + lim (ln (x)) = - \infty

= 0 - (- \infty)

Simplify

0 - (- \infty) : \infty

= \infty

\frac{d}{d x} ((x - 4)^{3})

f (x) = \frac{lo g _{9} ( x )}{7 + x ^{9}}

g (x) = 9 e^{x} x

\frac{d}{d x} (\frac{x ^{2}}{1 + 2 tan ( x )})

\frac{\partial}{\partial x} (\frac{x - y}{3 x y + 2})

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