Question

limit of (sqrt(x+1) -1)/ x as x approaches 0.

Answer 1

I think it would not exist.

Answer 2

\[\lim_{x \rightarrow 0}\frac{ \sqrt{x+1} -1}{ x }\]?

Answer 3

Would that mean you have to divide by the highest degree in the denominator?

Answer 4

If this is it then then multiply by the conjugate

Answer 5

Oh nevermind, Luigi is right.

Answer 6

Okay. I was gonna do that but i thought it was just for approaching infinity.

Answer 7

Yea! for the first time in my life I'm right :P

Answer 8

\[\lim_{x \rightarrow 0}\frac{ \sqrt{x+1} }{ x }*\frac{ \sqrt{x-1} }{ \sqrt{x-1} }\]

Answer 9

would it be 1?

Answer 10

and if you do it right then you should get an answer of 1/2

Answer 11

Bean you forgot the -1 D:

Answer 12

No, wouldnt be 1 i don't think. Depends on how you solve it?

Answer 13

yea! i know what i did wrong

Answer 14

Luigi there is no edit option D:

Answer 15

Haha that sucks. But anyways lucy do you understand?

Answer 16

Are you familiar with L'Hospital's or whatever that is

Answer 17

Yea i understand. but wouldn't the answer be DNE?

Answer 18

OMG! Nevermind!

Answer 19

You don't really need that here Sam.. I think

Answer 20

and no bean forgot that -1 at the end *cough*

Answer 21

LH rule applies when the limit goes to infinity and becomes indeterminant, correct?

Answer 22

it is 1/2, just wrote - instead of +

Answer 23

<_<

Answer 24

good job :P

Answer 25

I have one more i have trouble with. could you help me?

Answer 26

did you finish solving it?

Answer 27

and I gtg, so @Jhannybean has got you from here. Good luck! :P

Answer 28

\[\frac{ 1 }{ 3x ^{2} } - \frac{ 5x }{ x + 2 }\] as x approaches infinity.

Answer 29

I am not sure if i would have to have a common denominator or if it would be undefined.

Answer 30

I am going to make my best attempt at solving this problem :P

Answer 31

\[\lim_{x \rightarrow \infty}((\frac{ 1 }{ 3x^2 })-\frac{ 5x }{ x+2 })\] You can split the limit and solve each one separately\[\lim_{x \rightarrow \infty}\frac{ 1 }{ 3x^2 }- \lim_{x \rightarrow \infty}\frac{ 5x }{ x+2 }\]

Answer 32

the first would be negligible and the second would go to 5? so -5?

Answer 33

\[\frac{ 1 }{ 3 }\lim_{x \rightarrow \infty}x^2 - \lim_{x \rightarrow \infty}\frac{ \frac{ 5 }{ x } }{ \frac{ (x+2) }{x } }\]

Answer 34

oops i forgot the x of 5x! yes it would be 5

Answer 35

(5x/x)/[(x+2)/x]

Answer 36

\[\frac{ \sqrt{x+1} -1}{ x }\times \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\]
\[\frac{x}{x(\sqrt{x+1}+1)}\]
\[=\frac{1}{\sqrt{x+1}+1}\]

Answer 37

thanks! @Jhannybean

Answer 38

No problem! and Jim helped answer your previous question :P

Answer 39

wow i mean @satellite73