Mister Exam

Lang:

Limit of the function sqrt(1+n)/sqrt(2+n)

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     /  _______\
     |\/ 1 + n |
 lim |---------|
n->oo|  _______|
     \\/ 2 + n /

\lim_{n \to \infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right)

Limit(sqrt(1 + n)/sqrt(2 + n), n, oo, dir='-')

Lopital's rule

We have indeterminateness of type

oo/oo,

i.e. limit for the numerator is

\lim_{n \to \infty} \sqrt{n + 1} = \infty

and limit for the denominator is

\lim_{n \to \infty} \sqrt{n + 2} = \infty

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

\lim_{n \to \infty}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right)

\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \sqrt{n + 1}}{\frac{d}{d n} \sqrt{n + 2}}\right)

\lim_{n \to \infty}\left(\frac{\sqrt{n + 2}}{\sqrt{n + 1}}\right)

\lim_{n \to \infty}\left(\frac{\sqrt{n + 2}}{\sqrt{n + 1}}\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 n→0, -oo, +oo, 1

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

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

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

\lim_{n \to 1^-}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{6}}{3}

\lim_{n \to 1^+}\left(\frac{\sqrt{n + 1}}{\sqrt{n + 2}}\right) = \frac{\sqrt{6}}{3}

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

The graph