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 December 15th, 2018, 09:50 AM #1 Member   Joined: Feb 2018 From: Iran Posts: 56 Thanks: 3 Limit problem I tried to solve it this way But answer is not true where Did I go wrong? December 15th, 2018, 12:16 PM #2 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timey-wimey stuff. 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 Thanks from Elize December 15th, 2018, 01:00 PM   #3
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Quote:
 Originally Posted by Elize  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.

Last edited by JeffM1; December 15th, 2018 at 01:03 PM. December 15th, 2018, 01:28 PM #4 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 $x^2+5\approx x^2$ for large $|x|$. Therefore $\frac{x^3}{x^2+5}\approx \frac{x^3}{x^2}=x \to -\infty$, as $x\to -\infty$.. Thanks from greg1313 and topsquark December 16th, 2018, 04:30 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra \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} Thanks from topsquark December 16th, 2018, 07:34 AM   #6
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 Originally Posted by v8archie \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. December 16th, 2018, 10:25 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra If any teacher gave a F for that, I'd tell him to get over himself. I never divided by zero anyway. Tags limit, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jraed8 Calculus 2 December 14th, 2015 07:27 PM DysserDorf Calculus 8 December 6th, 2014 06:49 AM MFP Calculus 3 July 22nd, 2013 01:50 PM wawar05 Calculus 3 December 6th, 2011 02:14 AM peka0027 Real Analysis 1 February 25th, 2009 04:13 PM

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