Question

lim of x / ( sqrt( x + 1 ) - 1 = 2
x -> 0

x approaching to 0, the limit is 2 how

Answer 1

\[\LARGE \lim_{x\to 0}\frac{x}{\sqrt{x+1}-1}=2\]

Answer 2

yah. why lim = 2 ? how

Answer 3

I'm confused too O_O... :(

Answer 4

omg... :D it's alright :D

Answer 5

put the value of x=0

Answer 6

experiment, then we don't get 2 never

Answer 7

@Kreshnik how do you make that appear large??

Answer 8

try and see what you get??

Answer 9

then we get 0

Answer 10

yes that is the answer.

Answer 11

no the limit is = 2 . i have answer already. i have a graphic application that also indicates limit to be 2

Answer 12

i am getting answer to your question 1/2 not 0

Answer 13

then ... i must see ,,, OO ... didn't realized the indeterminate form.

Answer 14

multiply both by conjugate and see what happens/

Answer 15

but i want the answer to be 2. :) :)

Answer 16

@experimentX \[\huge \lim_{x\to 0} \frac{\text{Use \huge }}{\text{=this}}\]

lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2
\[\lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2\]

\large lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2
\[\large \lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2\]

\Large lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2
\[\Large \lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2\]

\LARGE \lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2
\[\LARGE \lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2\]

\huge \lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2
\[\huge \lim_{x \to 0} \frac{x}{\sqrt{x+1}-1}=2\]
@experimentX

Answer 17

yes ... multiplying by conjugate will give limit 2

Answer 18

yah experiment u r right ...

Answer 19

but applying L'hopitals give 1/2 :s

Answer 20

using L'hospital rule will also give the same 2 ... you might be forgetting something.

Answer 21

@experimentX please tell me how do you use L'H'R here.. both sides? :O :$

Answer 22

differentiate both function ... on top and bottom

Answer 23

experiment, how did u know that we have to multiply by denominator's conjugate?

Answer 24

experience man ... thats one of standard tricks.

Answer 25

\[ \lim_{x \rightarrow 0} \sin x / x\]

Answer 26

i need it's limit too

Answer 27

\[ \huge \lim_{x \to 0} \frac{\frac{d(x)}{dx}}{\frac{d(\sqrt{x+1}-1)}{dx}}=2 \]
\[ \huge \lim_{x \to 0} \frac{1}{\frac{1}{2\sqrt{x+1}}-0}=2 \]

Answer 28

hahaaaahh .. NOW we're talking ;) well done !

Answer 29

@Kreshnik put the value of x=0 on the top ... that is LHR
@jatinbansalhot it's value is 1 .... it's quite complicated => one is geometrical proof
other proof is quite complicated.

Answer 30

i have a new question plz try that one i urgently need that

Answer 31

@jatinbansalhot the best thing you can do to view it's proof is view MIT video lectures.

Answer 32

i had posted that

Answer 33

Experiment which lectures single variable calculus

Answer 34

single variable calculus is calculus 1 ?

Answer 35

send me exact link of those lectures plz .

Answer 36

i am beginner ... there's now way i could do that. @jatinbansalhot yes ... on single variable calculus. it's beautifully done there ... geometrically.

Answer 37

i'm doing Calculus 1 . Is single variable calculus same ?

Answer 38

yes ... lecture 3

Answer 39

only lecture 3 ? plz send me the link

Answer 40

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-3-derivatives/