Question

how can I delete indetermination in the following expression.

Limx(x->0)((sqrt(x+2)-sqrt(2))/x)

Answer 1

\[limit _{x->0}(\sqrt{x+2}-\sqrt{2})/2\] this is the equation

Answer 2

conjugation

Answer 3

limitx−>0(x+2−−−−−√−2−−√)/x Sorry

Answer 4

you mean by multiplyting the numerator and the denominator by \[\sqrt{x+2}+\sqrt{2}\]

Answer 5

Top and bottom

Answer 6

ok let me try

Answer 7

Changuanas, I actually get rid of indetermination. However when I apply the limit x->0, the answer is not the corrrect. The answer int he book is 1/2sqrt(2).

Answer 8

Sorry. I was wrong. I already got the answer! Can you help me with another exercise? Just to check

Answer 9

just post question, if fall asleep someone else would help

Answer 10

\[Limit _{x->infinity} \sqrt{3x ^{4}-2x+4}+x\]

Answer 11

I don't remember what happens to infinity under square root. Does it go to infinity?

Answer 12

According to the graph, it increases with no bound

Answer 13

Ignore all others and consider only highest exponent of x. What is highest exp?

Answer 14

I'm sorry. Did you already have an answer: infinity? or it needs more work?

Answer 15

yes, more work :)

Answer 16

I think the answer is 1

Answer 17

what is highest exponent x in your problem?

Answer 18

inside the root? its 2

Answer 19

Ignore all others divide sq rt 3x^4 by sq rt x^4

Answer 20

Wouldn't that one have to be just infinity?

Answer 21

I think he is in an early class. The class before you study infinity, right Emun?

Answer 22

yes

Answer 23

In that case, don't ignore everything else, that is what you do in higher math;keep everything and divide each term by x^2

Answer 24

It's funny being all done finals but still here during my break lol. why am I here?

Answer 25

except that expression isn't anything you would need to do that for. maybe if you were dividing by a polynomial

Answer 26

You here to help Emun I'm going to sleep.

Answer 27

guys the equation was wrong. The equation is the following:
\[\sqrt{x ^{2}-2x+4}+x\] I factored the expression inside the root out and I got
\[\sqrt{(x-2)^{2}}+x\] so finally I got
\[x-2+x\]
as I apply the limit as x=infinity,the answer I got is 2

Answer 28

Thanks changaunas :)

Answer 29

The answer would be infinity, wouldn't it?

Answer 30

wouldn't you get 2x-2 , which as x becomes large.. your expression does too?
infinity!

Answer 31

Yeah and this is the answer: Thanks gusy

Answer 32

thanks guys :)

Answer 33

\[\lim_{x \rightarrow 0} (\sqrt{x+2}-\sqrt{2})/x \times (\sqrt{x+2}+\sqrt{2})/(\sqrt{x+2}+\sqrt{2})\]
\[\lim_{x \rightarrow 0} (x+2-2)/(x(\sqrt{x+2}+\sqrt{2}))\]
\[\lim_{x \rightarrow 0} x/(x(\sqrt{x+2}+\sqrt{2}))\]
\[\lim_{x \rightarrow 0} 1/(\sqrt{x+2}+\sqrt{2})\]
\[1/(\sqrt{0+2}+\sqrt{2})\]
\[1/(\sqrt{2}+\sqrt{2})\]
\[1/(2\sqrt{2})\]