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Limit of the function (1+log(x))/x

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      /1 + log(x)\
 lim  |----------|
x->-oo\    x     /

\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} + 1}{x}\right)

Limit((1 + log(x))/x, x, -oo)

Lopital's rule

We have indeterminateness of type

oo/-oo,

i.e. limit for the numerator is

\lim_{x \to -\infty}\left(\log{\left(x \right)} + 1\right) = \infty

and limit for the denominator is

\lim_{x \to -\infty} x = -\infty

Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.

\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} + 1}{x}\right)

\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} \left(\log{\left(x \right)} + 1\right)}{\frac{d}{d x} x}\right)

\lim_{x \to -\infty} \frac{1}{x}

\lim_{x \to -\infty} \frac{1}{x}

0

It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)

The graph

Plot the graph

Rapid solution [src]

0

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} + 1}{x}\right) = 0

\lim_{x \to \infty}\left(\frac{\log{\left(x \right)} + 1}{x}\right) = 0

\lim_{x \to 0^-}\left(\frac{\log{\left(x \right)} + 1}{x}\right) = \infty

\lim_{x \to 0^+}\left(\frac{\log{\left(x \right)} + 1}{x}\right) = -\infty

\lim_{x \to 1^-}\left(\frac{\log{\left(x \right)} + 1}{x}\right) = 1

\lim_{x \to 1^+}\left(\frac{\log{\left(x \right)} + 1}{x}\right) = 1

The graph