Limit problem

Feb 2018
60
3
Iran

I tried to solve it this way

But answer is not true where Did I go wrong?
 

topsquark

Math Team
May 2013
2,385
996
The Astral plane
Who says that \(\displaystyle 0 \times \infty = 0\)? This is an indeterminate form.

Try it this way:
\(\displaystyle \lim _{x \to -\infty} \dfrac{x^3}{x^2 + 5}\)

\(\displaystyle = \lim _{x \to -\infty} \dfrac{x^3}{x^2 + 5} = \lim_{x \to -\infty} \left ( x - 5 + \dfrac{25x}{x^2 + 5} \right )\) by long division.

I leave it to you to prove \(\displaystyle \lim_{x \to -\infty} \dfrac{25x}{x^2 + 5} = 0\). So
\(\displaystyle \lim_{x \to -\infty} \dfrac{x^3}{x^2 + 5} = \lim_{x \to -\infty} (x - 5)\)

Can you fill in the steps?

-Dan
 
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May 2016
1,310
551
USA

I tried to solve it this way

But answer is not true where Did I go wrong?
I cannot read your jpeg clearly, but thank you for providing your work. As far as I can tell based on my poor reading of the jpeg, you made at least one error.

You cannot assume that $\infty * 0 = 0.$

Here is one way to attack the problem:

$\dfrac{x^3}{x^2 + 5} = \dfrac{x^3 + 5x^2 - 5x^2}{x^2 + 5} = \dfrac{x(x^2 + 5) - 5x^2}{x^2 + 5} = x + \dfrac{-\ 5x^2}{x^2 + 5} =$

$x + \dfrac{-\ 5x^2 - 25 + 25}{x^2 + 5} = x + \dfrac{-\ 5(x^2 + 5) + 25}{x^2 + 5} = x - 5 + \dfrac{25}{x^2 + 5}.$

With me so far? (You can get the same result using synthetic or long division.)

One of the laws of limits, speaking a bit loosely, is that the limit of a sum equals the sum of the limits.

$\displaystyle \therefore \lim_{x \rightarrow - \infty} \left ( \dfrac{x^3}{x^2 + 5} \right ) = \lim_{x \rightarrow - \infty} \left (x - 5 + \dfrac{25}{x^2 + 5} \right ) =$

$\displaystyle \lim_{x \rightarrow - \infty} (x) + \lim_{x \rightarrow -\infty} (-\ 5) + \lim_{x \rightarrow - \infty} \left ( \dfrac{25}{x^2 + 5} \right ) = - \ \infty - 5 + 0 = -\ \infty.$

Make sense?

EDIT: I see that Dan beat me to an answer.
 
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mathman

Forum Staff
May 2007
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$x^2+5\approx x^2$ for large $|x|$. Therefore $\frac{x^3}{x^2+5}\approx \frac{x^3}{x^2}=x \to -\infty$, as $x\to -\infty$..
 
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v8archie

Math Team
Dec 2013
7,709
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Colombia
\begin{align}
\lim_{x \to -\infty} \frac{x^3}{x^2+5} &= \lim_{y \to \infty} \frac{-y^3}{y^2+5} &\text{where $y=-x$} \\
&= \lim_{y \to \infty} \frac{\frac{-y^3}{y^3}}{\frac{y^2}{y^3}+\frac{5}{y^3}} \\
&= \lim_{y \to \infty} \frac{-1}{0+0} \\ &= \lim_{y \to \infty} \frac{-1}{0} \\ &= -\infty
\end{align}
 
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Oct 2009
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\begin{align}
\lim_{x \to -\infty} \frac{x^3}{x^2+5} &= \lim_{y \to \infty} \frac{-y^3}{y^2+5} &\text{where $y=-x$} \\
&= \lim_{y \to \infty} \frac{\frac{-y^3}{y^3}}{\frac{y^2}{y^3}+\frac{5}{y^3}} \\
&= \lim_{y \to \infty} \frac{-1}{0+0} \\ &= \lim_{y \to \infty} \frac{-1}{0} \\ &= -\infty
\end{align}
If one of my students wrote this down in my class, I'd give him an F on the spot. You can not and should never divide by 0.
 

v8archie

Math Team
Dec 2013
7,709
2,677
Colombia
If any teacher gave a F for that, I'd tell him to get over himself.

I never divided by zero anyway.
 
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