Results 1 to 4 of 4
Like Tree5Thanks
  • 3 Post By Plato
  • 2 Post By Prove It

Thread: I don't understand why the limit is negative infinity

  1. #1
    ORT
    ORT is offline
    Newbie
    Joined
    Jan 2019
    From
    Germany
    Posts
    5

    I don't understand why the limit is negative infinity

    Hello everyone,

    I have this task:
    I don't understand why the limit is negative infinity-screenshot-7-.png

    I tried to solve it and in my solution, I get positive infinity when I approach from the left (which is correct, according to the answer) and positive infinity which is false when I approach from the right.

    That is the solution: I don't understand why the limit is negative infinity-screenshot-5-.png

    Can someone explain, how I get the results?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,405
    Thanks
    3296
    Awards
    1

    Re: I don't understand why the limit is negative infinity

    Quote Originally Posted by ORT View Post
    I have this task:
    Click image for larger version. 

Name:	Screenshot (7).png 
Views:	5 
Size:	12.8 KB 
ID:	39235
    I tried to solve it and in my solution, I get positive infinity when I approach from the left (which is correct, according to the answer) and positive infinity which is false when I approach from the right.
    That is the solution: Click image for larger version. 

Name:	Screenshot (5).png 
Views:	9 
Size:	21.7 KB 
ID:	39236 Can someone explain, how I get the results?
    The numerator $x^3-3x^2 \approx -4$ for each $x\approx 2$ regardless of right or left.
    But the denominator $x-2<0$ for $x<2$ while $x-2>0$ for $x>2$
    Last edited by Plato; Jan 30th 2019 at 09:21 AM.
    Thanks from HallsofIvy, topsquark and ORT
    Follow Math Help Forum on Facebook and Google+

  3. #3
    ORT
    ORT is offline
    Newbie
    Joined
    Jan 2019
    From
    Germany
    Posts
    5

    Re: I don't understand why the limit is negative infinity

    Thank you very much!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,883
    Thanks
    1948

    Re: I don't understand why the limit is negative infinity

    Quote Originally Posted by ORT View Post
    Hello everyone,

    I have this task:
    Click image for larger version. 

Name:	Screenshot (7).png 
Views:	5 
Size:	12.8 KB 
ID:	39235

    I tried to solve it and in my solution, I get positive infinity when I approach from the left (which is correct, according to the answer) and positive infinity which is false when I approach from the right.

    That is the solution: Click image for larger version. 

Name:	Screenshot (5).png 
Views:	9 
Size:	21.7 KB 
ID:	39236

    Can someone explain, how I get the results?
    $\displaystyle \begin{align*} \lim_{x \to 2} \frac{x^3 - 3\,x^2}{x - 2} &= \lim_{x \to 2} \left( \frac{x^3 - 2\,x^2 - x^2}{ x - 2 } \right) \\ &= \lim_{x \to 2} \left( \frac{x^3 - 2\,x^2}{x - 2} - \frac{x^2}{x - 2} \right) \\ &= \lim_{x \to 2} \left[ \frac{x^2 \left( x - 2 \right) }{x - 2} - \frac{x^2}{x- 2} \right] \\ &= \lim_{x \to 2} \left( x^2 - \frac{x^2}{x - 2} \right) \\ &= \lim_{x \to 2} \left( x^2 + \frac{-x^2 + 2\,x - 2\,x}{x - 2} \right) \\ &= \lim_{x \to 2} \left[ x^2 + \frac{-x\left( x - 2 \right) - 2\,x}{x - 2} \right] \\ &= \lim_{x \to 2} \left[ x^2 - \frac{x \left( x - 2 \right) }{x - 2 } - \frac{2\,x}{x - 2} \right] \\ &= \lim_{x \to 2} \left( x^2 - x - \frac{2\,x}{x - 2} \right) \\ &= \lim_{x \to 2} \left( x^2 - x + \frac{-2\,x + 4 - 4}{x - 2} \right) \\ &= \lim_{x \to 2} \left[ x^2 - x + \frac{-2\,x + 4}{x - 2} - \frac{4}{x - 2} \right] \\ &= \lim_{x \to 2} \left[ x^2 - x + \frac{-2 \left( x - 2 \right) }{x - 2} - \frac{4}{x - 2} \right] \\ &= \lim_{x \to 2} \left( x^2 - x - 2 - \frac{4}{x - 2} \right) \end{align*}$

    The $\displaystyle \begin{align*} x^2 - x - 2 \to 0 \end{align*}$ as $\displaystyle \begin{align*} x \to 2 \end{align*}$ so the only important term is $\displaystyle \begin{align*} -\frac{4}{x - 2} \end{align*}$. As $\displaystyle \begin{align*} x \to 2^- \end{align*}$ we have $\displaystyle \begin{align*} -\frac{4}{x - 2} \to +\infty \end{align*}$ and as $\displaystyle \begin{align*} x \to 2^+ \end{align*}$ we have $\displaystyle \begin{align*} -\frac{4}{x - 2} \to -\infty \end{align*}$.

    Thus the limit does not exist.
    Thanks from topsquark and ORT
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit at negative infinity?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Oct 16th 2015, 03:50 AM
  2. Limit to negative infinity question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Feb 26th 2010, 04:51 AM
  3. limit to negative infinity
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: Feb 10th 2010, 02:45 AM
  4. limit as t goes to negative infinity...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 25th 2008, 12:22 PM
  5. limit to NEGATIVE infinity
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Nov 21st 2007, 03:36 PM

Search Tags


/mathhelpforum @mathhelpforum