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
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
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?
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.
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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