how can I delete indetermination in the following expression.
Limx(x->0)((sqrt(x+2)-sqrt(2))/x)

- anonymous

how can I delete indetermination in the following expression.
Limx(x->0)((sqrt(x+2)-sqrt(2))/x)

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- anonymous

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

- anonymous

conjugation

- anonymous

limitx−>0(x+2−−−−−√−2−−√)/x Sorry

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## More answers

- anonymous

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

- anonymous

Top and bottom

- anonymous

ok let me try

- anonymous

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).

- anonymous

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

- anonymous

just post question, if fall asleep someone else would help

- anonymous

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

- anonymous

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

- anonymous

According to the graph, it increases with no bound

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- anonymous

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

- anonymous

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

- anonymous

yes, more work :)

- anonymous

I think the answer is 1

- anonymous

what is highest exponent x in your problem?

- anonymous

inside the root? its 2

- anonymous

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

- anonymous

Wouldn't that one have to be just infinity?

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Thanks changaunas :)

- anonymous

The answer would be infinity, wouldn't it?

- anonymous

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

- anonymous

Yeah and this is the answer: Thanks gusy

- anonymous

thanks guys :)

- anonymous

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

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