0
$\begingroup$

I did:

$$\lim_{x \rightarrow 3} \frac{2x-e^{x-3}-5}{3-x} = \frac{-e^{x-3}+(-5+2x)}{-(x-3)} = \frac{-(e^{x-3}-(-5+2x))}{-(x-3)} = \frac{-(e^{y}-(-5+2x))}{-(y)} = \frac{e^{y}-(-5+2x)}{y}$$

Since $\lim_{x \rightarrow 3} -5+2x = 1$ and so $\lim_{x \rightarrow 3} y = 0$

$$\lim_{x \rightarrow 3}\frac{e^{y}-(-5+2x)}{y} = \frac{e^{y}-1}{y} = 1 $$

But the right answer is -1. I can't see where I swapped the sign though. What did I do wrong?

  • 0
    When x goes to 3,it's by the left, so the sign is negative. – 2017-02-24
  • 0
    And something smaller than 3 minus 3 is always negative. – 2017-02-24
  • 0
    If 0/0=1 (I'm guessing that's what you've assumed in your very last step, which is incorrect), then why even bother doing this whole procedure - you could've applied the same "logic" to the original expression of the limit. – 2017-02-24

3 Answers 3

1

You can write the original expression as $$\frac{e^{x-3} - 1}{x-3} + \frac{6 - 2x}{x-3} = \frac{e^{x-3} - 1}{x-3} - 2.$$

To compute the limit observe $$\lim_{x \to 3}\frac{e^{x-3} - 1}{x-3} = \left. \frac{d}{dx} e^{x-3} \right|_{x=3} = \left. e^{x-3}\right|_{x=3} = 1.$$

7

If you're allowed to use Taylor expansions:

$$e^{x-3}=1+(x-3)+\frac12(x-3)^2+\mathcal O((x-3)^3)$$

Thus, the limit reduces as follows:

$$\begin{align}\frac{2x-e^{x-3}-5}{3-x}&=\frac{(x-3)-\frac12(x-3)^2+\mathcal O((x-3)^3)}{-(x-3)}\\&=-1+\frac12(x-3)^2+\mathcal O((x-3)^2)\\&\to\boxed{-1}\end{align}$$

  • 0
    I think taylor expansion is the most intuitive way to do it. – 2017-02-24
  • 0
    Yes, as do I, and if anyone complains: http://math.stackexchange.com/tags/limits-without-lhopital/info – 2017-02-24
  • 1
    Don't worry people here normally and rightfully complain for L'Hospital's Rule and not Taylor. +1 – 2017-02-24
  • 0
    Correct grammar; it's not "your", it's "you're"! :-). – 2017-02-24
5

Let us set $x-3=y$.

Note that you cannot directly put $\lim\limits_{x \to 3}(-5+2x)=1$ here. This is like saying that since $\lim\limits_{x \to 0} x=0$, $$\lim_{x \to 0} \frac{x}{x}=\lim_{x \to 0} \frac{0}{x}=0$$ So, we can conclude; $$\lim_{x \to 3} \frac{2x-e^{x-3}-5}{3-x} \neq \lim_{y \to 0} \frac{e^y-1}{y}$$ Instead, as $-2x+5=-2x+6-1=-2y-1$, do $$\lim_{x \to 3} \frac{2x-e^{x-3}-5}{3-x} =\lim_{y \to 0} \frac{e^y-2y-1}{y}=\lim_{y \to 0}\frac{e^y-1}{y}-2=-1$$ For the future, I recommend that you try not to mix $x$ and $y$ up. It gets a bit confusing.

  • 0
    I think you should also change limit as $y\to 0$. – 2017-02-24
  • 0
    @ParamanandSingh Good point. – 2017-02-24
  • 0
    @ParamanandSingh I think I fixed it. Is it better now? – 2017-02-24
  • 0
    Hm, perhaps this will help?$$\frac{e^y-1}y-2$$ – 2017-02-24
  • 0
    @SimplyBeautifulArt Yes, I know this. I was recommending the OP to continue from here. – 2017-02-24
  • 0
    I don't think my edit was incorrect. In fact you have retained my edit. I had changed $+1$ to $-1$ in the last fraction numerator. – 2017-02-24
  • 0
    @ParamanandSingh Oh shoot. You did. Sorry. – 2017-02-24
  • 2
    No issues. Small things like these happen on big places like MSE. :) – 2017-02-24