2
$\begingroup$

During studying limits I regularly come across some of them, that cannot be simplified anymore using standard laws and formulas (e.g. substitutions, sum/product rules, L'Hôpital, etc.). However when the final expression is received it is still not obvious, how the function (resp. series) converges at a certain limit.

For instance $$ \lim_{s \to 0} {e^s-1 \over s} = 1 \;\;, \;\; \lim_{x \to \infty}e^{\log x \over x} = 1 $$

I somehow instinctively understand, that I should pay attention to how functions behave, which of them grows (resp. falls) faster than another. However I would like to know if there're some solid methods of analyzing such cases.

$\endgroup$
3
  • $\begingroup$ I would not call $\lim_{s \to 0} {e^s-1 \over s} = 1$ a trick. Since it means that the derivative of $exp$ at $0$ is $1$, this is just a part of a possible definition of the exponential function. Or this is the best trick ever: know your definitions... $\endgroup$
    – Taladris
    Jun 4, 2014 at 8:43
  • $\begingroup$ Joke aside, the second limit is equivalent to $\displaystyle{ \lim_{x\to\infty}\frac{ln(x)}{x}=0}$, that can be proved using L'Hospital rule. $\endgroup$
    – Taladris
    Jun 4, 2014 at 8:45
  • 1
    $\begingroup$ In this 13 February 2000 sci.math post archived at Math Forum, I posted several methods for evaluating limits (not going beyond the first two semesters of elementary calculus). This was taken from a longer handwritten class handout I used in the 1990s, which to this day I have not yet converted to a digital LaTeX document (but probably should get around to doing so at some point). $\endgroup$ Jun 4, 2014 at 14:13

3 Answers 3

1
$\begingroup$

If you're willing to accept that $\sum_{n=1}^\infty \frac{x^{n-1}}{n!}$ is uniformly convergent then you can use this to evaluate the first limit $$ \lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \sum_{n = 1}^\infty \frac{x^{n-1}}{n!} = \sum_{n=1}^\infty \frac{0^{n-1}}{n!} = 1 $$ In general for problems like these you can use power series to evaluate the limits. In other words if you know the power series representation of a function you can plug it into the limit and do some manipulation there.

For the second series I would take advantage of the continuity of $e^x$ and prove that $$ \lim_{x \to \infty} x\sqrt[x]{e} = 1 $$ which we can see $$ \lim_{x \to \infty} x\sqrt[x]{e} = \lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{x}{\sum_{n=0}^\infty \frac{x^n}{n!}} = \lim_{x \to \infty} \frac{1}{\frac{1}{x} + \sum_{n=1}^\infty \frac{x^{n-1}}{n!}} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x} + \sum_{n=2}^\infty \frac{x^{n-1}}{n!}} = 1 $$

Generally I like looking at power series to solve these limits.

Just thought of the somewhat traditional way to prove your first limit. Define $e = \lim_{n \to \infty} (1 + 1/n)^{n}$ and notice $$ \lim_{n \to \infty} (1 + 1/n)^n = \lim_{h \to 0} (1 + h)^{1/h} $$ so then $$ \lim_{s \to 0} \frac{e^s - 1}{s} = \lim_{s \to 0} \frac{\lim_{h \to 0} (1+h)^{s/h}-1}{s} $$ Now take advantage of $e^x$'s continuity to combine the limits $$ \lim_{s \to 0} \frac{(1+s)^{s/s}-1}{s} = 1 $$

$\endgroup$
1
  • $\begingroup$ That's useful, thanks $\endgroup$ Jun 4, 2014 at 9:29
1
$\begingroup$

Yes, use the fact that $$e^s = 1+ s + O(s^2).$$

Then we get $$\frac{e^s - 1}{s} = \frac{s + O(s^2)}{s} \to 1.$$

Alternatively note that the limit is equivalent to computing,

$$ \lim_{x \to 0} \frac{e^s - 1}{s} = \lim_{x \to 0} \frac{e^s - e^0}{s - 0} = \left (e^s)'\right|_{s=0} = e^0 = 1.$$

$\endgroup$
9
  • $\begingroup$ The second trick seems in some sense to be cheating. Unless I misunderstood your logic, your using the fact that you know that the derivative of $e^x$ is $e^x$ to compute the limit, when infact to show that the derivative of $e^x$ is shown to be $e^x$ by evaluating that limit. $\endgroup$
    – DanZimm
    Jun 4, 2014 at 8:57
  • $\begingroup$ If I swap the middle limits, does that help? I am trying to say that if you observe $1 = e^0$ and $s = s - 0$, and recalling the definition of a derivative, this is what we get. $\endgroup$
    – IAmNoOne
    Jun 4, 2014 at 8:59
  • 1
    $\begingroup$ Ok I observe that $s-0 = s, 1=e^0$, so I accept that $\lim_{s \to 0} \frac{e^s - e^0}{s-0} = \lim_{s \to 0} \frac{e^s - 1}{s}$ but this doesn't eliminate the circular logic. $\endgroup$
    – DanZimm
    Jun 4, 2014 at 9:01
  • 1
    $\begingroup$ @DanZimm I got your point, that's true, the derivative is defined itself as $\lim_{x \to 0} {f(x)-f(x_0) \over x-x_0}$ Thus there's indeed a circular logic in that explanation. $\endgroup$ Jun 4, 2014 at 9:08
  • 1
    $\begingroup$ Yes I understand, but how do you know that $(e^s)' = e^s$? $\endgroup$
    – DanZimm
    Jun 4, 2014 at 9:08
0
$\begingroup$

I think there aren't any methods that would cover every case,for example in this case $e^0=1$,and by the continuity of the exponential function as $s\to 0,e^s\to 1$ so from that $e^s-1\to 0$ and by the limit $s\to 0$ so it's 1.Also the above method with $O(s)$ is useful.Now for the second limit $x>\log(x)$ or $e^x>x$ since $e^x$ grows much faster than $x$ we can say that $\frac{\log(x)}{x}\to0$ as $x\to\infty$ so again since $e$ is continuos we say that limit is $1$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .