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December 15th, 2018, 09:50 AM   #1
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Unhappy Limit problem


I tried to solve it this way
But answer is not true where Did I go wrong?
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December 15th, 2018, 12:16 PM   #2
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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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December 15th, 2018, 01:00 PM   #3
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Quote:
Originally Posted by Elize View Post

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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Last edited by JeffM1; December 15th, 2018 at 01:03 PM.
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December 15th, 2018, 01:28 PM   #4
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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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December 16th, 2018, 04:30 AM   #5
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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}
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December 16th, 2018, 07:34 AM   #6
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Quote:
Originally Posted by v8archie View Post
\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.
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December 16th, 2018, 10:25 AM   #7
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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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