Asked by Anonymous
can someone please explain why the answer to this is negative infinity? I keep getting positive.
lim
x--> - infinity x^3-2/x^2+x
lim
x--> - infinity x^3-2/x^2+x
Answers
Answered by
Bosnian
Go on: wolframalpha dot com
When page be open in rectangle type:
limit x^3-2/x^2+x as x-> - infinity
and click option =
After few seconds when you see result click option:
Show steps
On wolfram alpha dot com you can practice any kind of calculus.
That is good just for practice.
You can't use wolframalpha on exam.
When page be open in rectangle type:
limit x^3-2/x^2+x as x-> - infinity
and click option =
After few seconds when you see result click option:
Show steps
On wolfram alpha dot com you can practice any kind of calculus.
That is good just for practice.
You can't use wolframalpha on exam.
Answered by
Reiny
Just get a feel for the numbers
x ---> - infinity
for very large negative numbers , x^3 - 2 becomes "
"hugely negative"
but x^2 + x becomes + "very large"
since -/+ = - , and the numerator is larger than the denominator by a factor of x,
the answer is -negative infinity
or
(x^3 - 2)/(x^2 + x) = x - 1 + (x+2)/(x^2+x)
lim (x^3-2)/(x^2+x) as x---> -∞
= lim (x-1) + lim (x+2)/(x^2+x) as x---> -∞
since intuitively we can see that lim (x+2)/(x^2+x) --> 0 as x ---> -∞
we are left with lim x-1 as x -->-∞
which is -∞
x ---> - infinity
for very large negative numbers , x^3 - 2 becomes "
"hugely negative"
but x^2 + x becomes + "very large"
since -/+ = - , and the numerator is larger than the denominator by a factor of x,
the answer is -negative infinity
or
(x^3 - 2)/(x^2 + x) = x - 1 + (x+2)/(x^2+x)
lim (x^3-2)/(x^2+x) as x---> -∞
= lim (x-1) + lim (x+2)/(x^2+x) as x---> -∞
since intuitively we can see that lim (x+2)/(x^2+x) --> 0 as x ---> -∞
we are left with lim x-1 as x -->-∞
which is -∞
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