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

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

Mathematics
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\[limit _{x->0}(\sqrt{x+2}-\sqrt{2})/2\] this is the equation
conjugation
limitx−>0(x+2−−−−−√−2−−√)/x Sorry

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Other answers:

you mean by multiplyting the numerator and the denominator by \[\sqrt{x+2}+\sqrt{2}\]
Top and bottom
ok let me try
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).
Sorry. I was wrong. I already got the answer! Can you help me with another exercise? Just to check
just post question, if fall asleep someone else would help
\[Limit _{x->infinity} \sqrt{3x ^{4}-2x+4}+x\]
I don't remember what happens to infinity under square root. Does it go to infinity?
According to the graph, it increases with no bound
1 Attachment
Ignore all others and consider only highest exponent of x. What is highest exp?
I'm sorry. Did you already have an answer: infinity? or it needs more work?
yes, more work :)
I think the answer is 1
what is highest exponent x in your problem?
inside the root? its 2
Ignore all others divide sq rt 3x^4 by sq rt x^4
Wouldn't that one have to be just infinity?
I think he is in an early class. The class before you study infinity, right Emun?
yes
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
It's funny being all done finals but still here during my break lol. why am I here?
except that expression isn't anything you would need to do that for. maybe if you were dividing by a polynomial
You here to help Emun I'm going to sleep.
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
Thanks changaunas :)
The answer would be infinity, wouldn't it?
wouldn't you get 2x-2 , which as x becomes large.. your expression does too? infinity!
Yeah and this is the answer: Thanks gusy
thanks guys :)
\[\lim_{x \rightarrow 0} (\sqrt{x+2}-\sqrt{2})/x \times (\sqrt{x+2}+\sqrt{2})/(\sqrt{x+2}+\sqrt{2})\] \[\lim_{x \rightarrow 0} (x+2-2)/(x(\sqrt{x+2}+\sqrt{2}))\] \[\lim_{x \rightarrow 0} x/(x(\sqrt{x+2}+\sqrt{2}))\] \[\lim_{x \rightarrow 0} 1/(\sqrt{x+2}+\sqrt{2})\] \[1/(\sqrt{0+2}+\sqrt{2})\] \[1/(\sqrt{2}+\sqrt{2})\] \[1/(2\sqrt{2})\]

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