Mister Exam

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

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

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

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

Lopital's rule

We have indeterminateness of type

oo/oo,

i.e. limit for the numerator is

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

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

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

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

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

\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{e^{x}}{2}}{\frac{d}{d x} x}\right)

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

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

\infty

It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)

The graph

Plot the graph

Other limits x→0, -oo, +oo, 1

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

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

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

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

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

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

Rapid solution [src]

oo

\infty

Expand and simplify

The graph