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x+log(x)/x

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

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     /    log(x)\
 lim |x + ------|
x->oo\      x   /
limx(x+log(x)x)\lim_{x \to \infty}\left(x + \frac{\log{\left(x \right)}}{x}\right)
Limit(x + log(x)/x, x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
oo/oo,

i.e. limit for the numerator is
limx(x2+log(x))=\lim_{x \to \infty}\left(x^{2} + \log{\left(x \right)}\right) = \infty
and limit for the denominator is
limxx=\lim_{x \to \infty} x = \infty
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
limx(x+log(x)x)\lim_{x \to \infty}\left(x + \frac{\log{\left(x \right)}}{x}\right)
=
Let's transform the function under the limit a few
limx(x2+log(x)x)\lim_{x \to \infty}\left(\frac{x^{2} + \log{\left(x \right)}}{x}\right)
=
limx(ddx(x2+log(x))ddxx)\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x^{2} + \log{\left(x \right)}\right)}{\frac{d}{d x} x}\right)
=
limx(2x+1x)\lim_{x \to \infty}\left(2 x + \frac{1}{x}\right)
=
limx(2x+1x)\lim_{x \to \infty}\left(2 x + \frac{1}{x}\right)
=
\infty
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
02468-8-6-4-2-1010-5050
Other limits x→0, -oo, +oo, 1
limx(x+log(x)x)=\lim_{x \to \infty}\left(x + \frac{\log{\left(x \right)}}{x}\right) = \infty
limx0(x+log(x)x)=\lim_{x \to 0^-}\left(x + \frac{\log{\left(x \right)}}{x}\right) = \infty
More at x→0 from the left
limx0+(x+log(x)x)=\lim_{x \to 0^+}\left(x + \frac{\log{\left(x \right)}}{x}\right) = -\infty
More at x→0 from the right
limx1(x+log(x)x)=1\lim_{x \to 1^-}\left(x + \frac{\log{\left(x \right)}}{x}\right) = 1
More at x→1 from the left
limx1+(x+log(x)x)=1\lim_{x \to 1^+}\left(x + \frac{\log{\left(x \right)}}{x}\right) = 1
More at x→1 from the right
limx(x+log(x)x)=\lim_{x \to -\infty}\left(x + \frac{\log{\left(x \right)}}{x}\right) = -\infty
More at x→-oo
Rapid solution [src]
oo
\infty
The graph
Limit of the function x+log(x)/x