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Limit of the function 2*x/(1+x^2)

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     / 2*x  \
 lim |------|
x->oo|     2|
     \1 + x /

\lim_{x \to \infty}\left(\frac{2 x}{x^{2} + 1}\right)

Limit((2*x)/(1 + x^2), x, oo, dir='-')

Detail solution

Let's take the limit

\lim_{x \to \infty}\left(\frac{2 x}{x^{2} + 1}\right)

Let's divide numerator and denominator by x^2:

\lim_{x \to \infty}\left(\frac{2 x}{x^{2} + 1}\right)

\lim_{x \to \infty}\left(\frac{2 \frac{1}{x}}{1 + \frac{1}{x^{2}}}\right)

Do Replacement

u = \frac{1}{x}

then

\lim_{x \to \infty}\left(\frac{2 \frac{1}{x}}{1 + \frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{2 u}{u^{2} + 1}\right)

\frac{0 \cdot 2}{0^{2} + 1} = 0

The final answer:

\lim_{x \to \infty}\left(\frac{2 x}{x^{2} + 1}\right) = 0

Lopital's rule

We have indeterminateness of type

oo/oo,

i.e. limit for the numerator is

\lim_{x \to \infty}\left(2 x\right) = \infty

and limit for the denominator is

\lim_{x \to \infty}\left(x^{2} + 1\right) = \infty

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

\lim_{x \to \infty}\left(\frac{2 x}{x^{2} + 1}\right)

=
Let's transform the function under the limit a few

\lim_{x \to \infty}\left(\frac{2 x}{x^{2} + 1}\right)

\lim_{x \to \infty}\left(\frac{\frac{d}{d x} 2 x}{\frac{d}{d x} \left(x^{2} + 1\right)}\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{2 x}{x^{2} + 1}\right) = 0

\lim_{x \to 0^-}\left(\frac{2 x}{x^{2} + 1}\right) = 0

\lim_{x \to 0^+}\left(\frac{2 x}{x^{2} + 1}\right) = 0

\lim_{x \to 1^-}\left(\frac{2 x}{x^{2} + 1}\right) = 1

\lim_{x \to 1^+}\left(\frac{2 x}{x^{2} + 1}\right) = 1

\lim_{x \to -\infty}\left(\frac{2 x}{x^{2} + 1}\right) = 0

The graph