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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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    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,2011-08-23
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    $(1+0)^0 \to 1^0 \to 1$, no indeterminancy here. The indeterminancy appears in limits of the form $1^\infty$, eg $\lim\limits_{x\to0}(1 + x)^{1/x}$2011-08-23
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    I asked the wrong question, I'm sorry. You were guessing for the right question.. could you provide me with a solution now?2011-08-23

5 Answers 5

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

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

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    With exp(x) do you mean $e^x$?2011-08-23
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    @Mats:$\quad$yes!2011-08-23
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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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    This is actually the best answer! You take a nice teaching approach!2011-08-23
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    @Mats: Thank you. Considering you said you had an exam tomorrow, I figured I'd explain the more general strategy :)2011-08-23
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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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    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.2011-08-23
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    @Mats: I added another approach. However, both yield the same answer, but in different forms.2011-08-23
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    @Mats: The first suggested solution is fine, it just requires some work. There is no universal "right solution".....2011-08-23
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    @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.2011-08-23
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    @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$2011-08-23
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    @Eric: In school there is a right solution :) If i won't use the second method he sugested, which seems fine but is better formulated in sunni's answer, I won't get any points.2011-08-23
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    @FoolForMath: It should be noted that to do this rigorously, you have to carefully manage the error.2011-08-23
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    @Mats: for the question asked, there is no right answer. Different approaches expose different aspects of the math involved. If you are looking for a particular answer, the question needs to be more focused, or you just need to expect to sift through a lot of different approaches. Saying to people, who have gone to some amount of work to answer what they think your question is, "that is not the right solution" is not the right response.2011-08-23
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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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    @AidenStrydom See last formula of your answer.2012-09-10
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    Dear sir - my apologies i miss read your answer. Please forgive me2012-09-10