Mister Exam

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Limit of the function sqrt(4+x^2)/x

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     /   ________\
     |  /      2 |
     |\/  4 + x  |
 lim |-----------|
x->oo\     x     /

\lim_{x \to \infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right)

Limit(sqrt(4 + x^2)/x, x, oo, dir='-')

Lopital's rule

We have indeterminateness of type

oo/oo,

i.e. limit for the numerator is

\lim_{x \to \infty} \sqrt{x^{2} + 4} = \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{\sqrt{x^{2} + 4}}{x}\right)

\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \sqrt{x^{2} + 4}}{\frac{d}{d x} x}\right)

\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 4}}\right)

\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 4}}\right)

1

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]

1

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to \infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = 1

\lim_{x \to 0^-}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = -\infty

\lim_{x \to 0^+}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \infty

\lim_{x \to 1^-}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \sqrt{5}

\lim_{x \to 1^+}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = \sqrt{5}

\lim_{x \to -\infty}\left(\frac{\sqrt{x^{2} + 4}}{x}\right) = -1

The graph