Asked by Anonymous

How do you solve a limit when it approaches infinity and there is a square root in the denominator? say the highest power is x^2 in both the numerator and denominator.

Answers

Answered by Reiny
as x ---> inf
and the top and bottom have the same highest power, then those terms will determine the limit

See my answer to a similar question by James at 11:06
Answered by Anonymous
I understand that problem. I get confused with the square root in the denominator. Aren't you supposed to do something with absolute values?
Answered by Reiny
did you mean something like

lim (4x^2 -2√x)/(3x^2 + 5√2x)
of course it could only approach + inf , or else the root terms would be undefined.
Other than that, the same rule applies
the limit would be 4/3
Answered by Anonymous
Yeah like
limit as x approaches infinity( 3x^2+2x)/(sqrt(x^2-2x)

Would it equal 3 then?
Answered by Reiny
No,
notice the highest powers are NOT the same, so my explanation is not valid.

as x --> inf
√(x^2 - 2x) ---> x
e.g. pick x = 1000000 and use your calculator to see

so the numerator ---> 3x^2
the denominator ----> x

so the limit is inf.
Answered by Anonymous
Ok I understand it now. Thank you.
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