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How would I work out a limit of the form:

$$\lim_{x\to 0}\;(1+x)^{1/x}$$

I know these types of limits have a solution based on $e$ but how do I find this solution?

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    $\begingroup$ The limit is $1$. $1^0$ is not an indeterminate form. If you asked for $\lim_{x\to 0} (1+x)^{1/x}$, that would be a different story, $\endgroup$ Commented Aug 23, 2011 at 20:43
  • $\begingroup$ $(1+0)^0 \to 1^0 \to 1$, no indeterminancy here. The indeterminancy appears in limits of the form $1^\infty$, eg $\lim\limits_{x\to0}(1 + x)^{1/x}$ $\endgroup$
    – leonbloy
    Commented Aug 23, 2011 at 20:43
  • $\begingroup$ I asked the wrong question, I'm sorry. You were guessing for the right question.. could you provide me with a solution now? $\endgroup$
    – Mats
    Commented Aug 23, 2011 at 20:48

5 Answers 5

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$\lim\limits_{x \rightarrow 0}\exp (x\ln (1+x))=\exp(\lim\limits_{x \rightarrow 0}(x\ln(1+x)))=\exp(0)=1$.

$\lim\limits_{ x \rightarrow 0}\exp ( \frac{\ln (1+x)}{x})=\exp(\lim\limits_{x \rightarrow 0}(\frac{\ln(1+x)}{x}))=\exp(1)=e$. Use L'Hospital.. to see $\lim\limits_{x \rightarrow 0}\frac{\ln(1+x)}{x}=1$

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  • $\begingroup$ With exp(x) do you mean $e^x$? $\endgroup$
    – Mats
    Commented Aug 23, 2011 at 21:01
  • $\begingroup$ @Mats:$\quad$yes! $\endgroup$
    – Quixotic
    Commented Aug 23, 2011 at 21:06
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Hint:

The functions $y = \log x$ and $y = e^x$ are continuous, and continuous functions respect limits: $$ \lim_{n \to \infty} f(g(n)) = f\left( \lim_{n \to \infty} g(n) \right), $$ for all continuous functions $f$, whenever $\displaystyle\lim_{n \to \infty} g(n)$ exists. Let $$L=\lim\limits_{x\to 0}(1+x)^{1/x}$$ be the limit which you to wish to find. Instead of finding $L$ directly, try on your own to find $\ln(L)$.

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  • $\begingroup$ This is actually the best answer! You take a nice teaching approach! $\endgroup$
    – Mats
    Commented Aug 23, 2011 at 21:24
  • $\begingroup$ @Mats: Thank you. Considering you said you had an exam tomorrow, I figured I'd explain the more general strategy :) $\endgroup$
    – JavaMan
    Commented Aug 23, 2011 at 23:51
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Try writing this as $$ \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n $$ The binomial theorem may be of some help then.

Another way of looking at this is taking logs and using L'Hospital $$ \log\left(\lim_{x\to0}(1+x)^{1/x}\right)=\lim_{x\to0}\frac{\log(1+x)}{x} $$

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  • $\begingroup$ This is not the right solution .. It was something with putting the whole limit as a power of $e$ .. but I forgot the correct solution method. $\endgroup$
    – Mats
    Commented Aug 23, 2011 at 20:54
  • $\begingroup$ @Mats: I added another approach. However, both yield the same answer, but in different forms. $\endgroup$
    – robjohn
    Commented Aug 23, 2011 at 20:56
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    $\begingroup$ @Mats: The first suggested solution is fine, it just requires some work. There is no universal "right solution"..... $\endgroup$ Commented Aug 23, 2011 at 20:59
  • $\begingroup$ @Eric: The Binomial Theorem can produce a series for the limit that converges pretty fast. The limit is usually the definition for $e$, and if that is the case, all that is left is to compute the value. $\endgroup$
    – robjohn
    Commented Aug 23, 2011 at 20:59
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    $\begingroup$ @Mats:This is a correct solution,put $\frac{1}{x} = n$ then as $x\to 0 \Rightarrow n\to\infty$ after this expand using multinomial/binomial theorem with rational index,arranging the terms and substituting the limit you will get this $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots = e$ $\endgroup$
    – Quixotic
    Commented Aug 23, 2011 at 21:02
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If we use the substitution $x=\frac{1}{y}$, since $\lim_{x\rightarrow 0}x=\lim_{y\rightarrow \infty }\frac{1}{y}$, we get

$$\lim_{x\rightarrow 0}\left( 1+x\right) ^{1/x}=\lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}=e,$$

which uses the result

$$\lim_{n\rightarrow \infty}\left( 1+\frac{1}{n}\right) ^{n}=e.$$

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    $\begingroup$ @AidenStrydom See last formula of your answer. $\endgroup$ Commented Sep 10, 2012 at 7:21
  • $\begingroup$ Dear sir - my apologies i miss read your answer. Please forgive me $\endgroup$ Commented Sep 10, 2012 at 8:09
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Sorry for uploading the image - i am new and have yet to figure out how to mark up the math

The first line assumes you know that if f(x) = ln(x) then f'(1) = (ln(x+h) - ln(x)) / h

ps f'(1) = 1 enter image description here

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