Question

Limit

Answer 1

\[\lim_{n \rightarrow \infty} n*\ln(\sqrt{n^2+2n+5}-n)\]

Answer 2

have you tried plugging in
x =1/n

as n-> infty
x->0

Answer 3

so...

Answer 4

I've applied L'hopital rule

Answer 5

you can use L'H right ?

Answer 6

do it after the substitution

Answer 7

if you plug it in it would be infinity - infinity ...
because the sqrt would be approaching infinity

Answer 8

yeah L'H's

Answer 9

rule *

Answer 10

L'H can be used only for
0/0 or inf/inf form
hence that substitution is required

Answer 11

dude I've already apllied l'hopital rule but the expression doesn't simplify at all
@hartnn I know the theory very well

Answer 12

\[\lim_{n \rightarrow \infty}\frac{ \ln(\sqrt{n^2+2n+5}-n) }{ \frac{ 1 }{ n } } \]
do you ever tried this?

Answer 13

\(\ln (5x^2 -2x+1) -\ln x \\ \dfrac{10x^2 -2}{5x^2 -2x+1} -\dfrac{1}{x}\)

you tried something like this ?

Answer 14

I know what to do.

\[\Large\color{blue}{ \bf \lim_{n \rightarrow infinity} \frac{\ln\sqrt{n^2+2x+5}}{\frac{1}{n}} }\]

Answer 15

I mean\[\Large\color{blue}{ \bf \lim_{n \rightarrow infinity} \frac{\ln\sqrt{n^2+2x+5}-n}{\frac{1}{n}} }\]

Answer 16

that would be ugly

Answer 17

10 points for zelman

Answer 18

as you plug infinity you get inf-inf on top
and 1/inf on bottom

Answer 19

then derive top bottom

Answer 20

on top is actually 0

Answer 21

0/0

Answer 22

BOTTOM IS ZERO

Answer 23

YES 0/0

Answer 24

yup, infinity - infinity

Answer 25

\[\Large\color{blue}{ \bf \lim_{n \rightarrow infinity} \frac{\ln\sqrt{n^2+2n+5}-n}{\frac{1}{n}} } \]

Answer 26

2n not 2x, srry