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I want to find the following limit:

$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$$

This is what I do. I change the variable $t=-x$ and I have the following limit:

$$\lim_{t\rightarrow +\infty}{e^{\frac {1}{2+t}}\cdot{\frac{t^2-2t-1}{-t-2}}+t}=\lim_{t\rightarrow+\infty}{h(x)}$$

We have $e^{\frac{1}{2+t}}\rightarrow1$ for $t\rightarrow+\infty$

Therefore I think (this is the passage I'm less sure about)

$$h(x)\sim \frac{t^2-2t-1}{-t-2}+t=\frac{t^2-2t-1+t(-t-2)}{-t-2}=\frac{t^2-2t-1-t^2-2t}{-t-2}=\frac{-4t-1}{-t-2}\sim{\frac {-4t}{-t}}\rightarrow4$$

The solution should actually be $3$. Any hints on what I'm doing wrong?

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  • $\begingroup$ You went wrong taking $e^{\frac{1}{2+t}}\rightarrow1$, we need to expand to one term more. $\endgroup$
    – user
    Sep 3, 2018 at 10:23
  • $\begingroup$ In general one can't replace a small part of a large expression by its limit while evaluating the limit of the large expression. Such replacements are allowed in specific cases and you may have a look at this answer to get more details on the specific cases. $\endgroup$
    – Paramanand Singh
    Sep 3, 2018 at 15:11

4 Answers 4

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ALternatively: $$\lim_{x\rightarrow -\infty}\left[{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x\right]=\\ \lim_{x\to-\infty} \left[e^{\frac{1}{2-x}} \left(x+4+\frac7{x-2}\right)-x\right]=\\ \lim_{x\to-\infty} \left[e^{\frac{1}{2-x}}\left(x+4\right)-x\right] +\underbrace{\lim_{x\to-\infty} e^{\frac{1}{2-x}} \cdot \frac7{x-2}}_{=0}=\\ \lim_{x\to-\infty} \left[e^{\frac{1}{2-x}}\left(x+4\right)-(x+4)+4\right]=\\ \lim_{x\to-\infty} \left[\frac{e^{\frac1{2-x}}-1}{\frac1{2-x}}\cdot \frac{x+4}{2-x}+4\right]=\\ 1\cdot (-1)+4=3,$$ where it was used: $$\lim_{x\to 0} \frac{e^x-1}{x}=1.$$

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  • $\begingroup$ Thanks, this looks clean. If I may ask how do you get $\frac{x^2+2x-1}{x-2}=(x+4+\frac{7}{x-2})$. When I try to find the solutions for $x^2+2x-1$ I get roots $\endgroup$
    – Cesare
    Sep 3, 2018 at 11:13
  • $\begingroup$ you can rearrange: $\frac{x^2-4+2x-4+7}{x-2}$ $\endgroup$
    – farruhota
    Sep 3, 2018 at 11:14
  • $\begingroup$ Somebody like me will definitely need a certain "eye" to do some of the things you did. There's certainly some good level of experience involved here. But I think this is by far the cleanest and most straightforward solution. Thanks $\endgroup$
    – Cesare
    Sep 3, 2018 at 11:30
  • $\begingroup$ You are welcome. There are several overlapping points with other solutions, so the credit goes to all. Good luck. $\endgroup$
    – farruhota
    Sep 3, 2018 at 11:36
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I think the first substitution isn't too useful, but let's keep it nontheless. $$e^{1/(2+x)}\frac{-x^2+2x+1}{x+2}+x=(4-x)e^{1/(2+x)}-7\frac{e^{1/(2+x)}}{x+2}+x=\\=4e^{1/(x+2)}-\frac7{x+2}e^{1/(2+x)}+xe^{1/(2+x)}\left(e^{-1/(2+x)}-1\right)=\\=4e^{1/(x+2)}-\frac7{x+2}e^{1/(2+x)}-\frac x{x+2}e^{1/(2+x)}\frac{e^{-1/(2+x)}-1}{-\frac1{2+x}}$$

Now, since $-\frac1{x+2}\to 0^-$ as $x+2\to\infty$ and $\lim_{s\to 0}\frac{e^s-1}s=1$, $$\lim_{x\to\infty}4e^{1/(x+2)}-\frac7{x+2}e^{1/(2+x)}-\frac x{x+2}e^{1/(2+x)}\frac{e^{-1/(2+x)}-1}{-\frac1{2+x}}=\\=\left[4e^0-\frac7\infty e^0-1\cdot e^0\cdot 1\right]=3$$

Now that you should know one more idea than before, you can see for yourself how your initial guess fares against $$\lim_{x\to\infty} xe^{1/(\alpha +\beta x)}-x,\qquad \beta\ne 0$$ as opposed to the correct answer.

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  • $\begingroup$ Thanks Saucy, honestly I don't understand most of this. I'm sure this is correct +1 $\endgroup$
    – Cesare
    Sep 3, 2018 at 10:45
  • $\begingroup$ You shouldn't trust it, because the first version had a mistake of algebra. Fortunately, in a term that then decays to $0$. $\endgroup$
    – user562983
    Sep 3, 2018 at 10:45
  • $\begingroup$ Are you using Taylor here? $\endgroup$
    – Cesare
    Sep 3, 2018 at 10:46
  • $\begingroup$ Not really. Just multiplication-and-division, a couple of fundamental limits and an eye for substituitions. $\endgroup$
    – user562983
    Sep 3, 2018 at 10:47
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Although $\lim\limits_{x\to-\infty}e^{\frac1{2-x}}=1$, we cannot simply replace it by $1$ in the limit because it is actually $1+\frac1{2-x}+O\!\left(\frac1{(2-x)^2}\right)$


For $x$ near $\pm\infty$, $$ \begin{align} e^{\frac1{2-x}}\frac{x^2+2x-1}{x-2}-x &=\overbrace{\left(1+\frac1{2-x}+O\!\left(\frac1{(2-x)^2}\right)\right)}^{e^{\frac1{2-x}}}\overbrace{\left(x+4-\frac7{2-x}\right)\vphantom{\left(\frac1{()^2}\right)}}^{\frac{x^2+2x-1}{x-2}}-x\\ &=\frac x{2-x}+4+O\!\left(\frac1{2-x}\right) \end{align} $$ Now let $x\to-\infty$.


Simply replacing $e^{\frac1{2-x}}$ by $1$ misses the $\frac x{2-x}$ in the formula above and leaves us only with $4$ in the limit.

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Let set $\frac {1}{2-x}=y \to 0$

$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x=\lim_{y\rightarrow 0}{e^{y}\cdot\frac{\frac1{y^2}-\frac 6 y+7}{-\frac1y}}-2+\frac1y$$$$=\lim_{y\rightarrow 0}e^{y}\cdot\left(-\frac1{y}+6-7y\right)-2+\frac1y$$

then use $e^y=1+y+o(y)$ to obtain

$$(1+y+o(y))\cdot\left(-\frac1{y}+6-7y\right)-2+\frac1y$$

$$-\frac1{y}+6-7y-1+6y-7y^2+o(1)-2+\frac1y=3+o(1) \to 3$$

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  • $\begingroup$ I don't understand how to replace $x^2$ with $\frac{1}{2-x}$. I tried to get $x$ out of $\frac{1}{2-x}$.... $\endgroup$
    – Cesare
    Sep 3, 2018 at 10:15
  • $\begingroup$ Do I need to find the inverse? Replacing $x$ with something that contains the variable $y$ isn't easy for me $\endgroup$
    – Cesare
    Sep 3, 2018 at 10:18
  • $\begingroup$ @Cesare I add some detail more. $\endgroup$
    – user
    Sep 3, 2018 at 10:19
  • $\begingroup$ Thanks, are you using Taylor? $\endgroup$
    – Cesare
    Sep 3, 2018 at 10:22
  • $\begingroup$ @Cesare Yes of course, but note that it always necessary tak einto account the remainder terms vie little-o or big-O notation otherwise we can easily get wrong. $\endgroup$
    – user
    Sep 3, 2018 at 10:24

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