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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 timeywimey 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 
December 15th, 2018, 01:00 PM  #3  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  Quote:
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 $x\to \infty$..

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} 
December 16th, 2018, 07:34 AM  #6  
Senior Member Joined: Oct 2009 Posts: 884 Thanks: 340  Quote:
 
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. 

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