anonymous
  • anonymous
how can I delete indetermination in the following expression. Limx(x->0)((sqrt(x+2)-sqrt(2))/x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[limit _{x->0}(\sqrt{x+2}-\sqrt{2})/2\] this is the equation
anonymous
  • anonymous
conjugation
anonymous
  • anonymous
limitx−>0(x+2−−−−−√−2−−√)/x Sorry

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