Question

Can someone please help me solve this limit?

Answer 1

where is it?

Answer 2

\[\large\lim_{n\to\infty}(\sqrt{2n+1}-\sqrt{n^2+3})\]

Answer 3

do you have options to choose from?

Answer 4

the second term will dominate the limit at infty

Answer 5

I was thinking of splitting the limit in 2, the second one is infty but i don't know what will the first one be.

Answer 6

maybe it will help to visualise a sqrt function

Answer 7

\[\lim_{n \rightarrow \infty} \sqrt{2n+1} - \lim_{n \rightarrow \infty} \sqrt{n ^{2}+3} \]

Answer 8

is it correct this way?

Answer 9

that way wont work this time, because you have to compare infinities

Answer 10

\[\lim_{n \rightarrow \infty} \sqrt{2n+1} - \lim_{n \rightarrow \infty} \sqrt{n ^{2}+3}\\
~\\
\leadsto \infty-\infty\\
=?\]

Answer 11

yep, not defined limit

Answer 12

should i use l'hopital?

Answer 13

if you realise before splitting the limit up that one of the terms dominates, then you can ignore the other term

Answer 14

can you use l'hop when the intermediate form is \(\infty\)?

i thought to use l'hop you need \(\frac\infty\infty\), or \(\frac00\), etc, with a fraction

Answer 15

you're right. sorry, it has been a while since we did this in highschool -.-

Answer 16

the sqrt n^2 term is dominant over the sqrt n, as n approaches infty

Answer 17

These \(\color{gray}{gray}\) terms do not need to be considered as n approaches infty
\[\large\lim_{n\to\infty}(\color{gray}{\sqrt{2n+1}}-\sqrt{n^2\color{gray}{+3}})\]

Answer 18

what do you think the limit will tend towards ?

Answer 19

\[\large\lim_{n\to\infty}(-\sqrt{n^2})\]

Answer 20

-inf?

Answer 21

that's what i get

Answer 22

it is probably -inf, but to show this analytically you can
multiply top and bottom by the conjugate