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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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    I asked the wrong question, I'm sorry. You were guessing $f$or 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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    @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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    @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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    @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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    Dear sir - my apologies i miss read your answer. Please forgive me2012-09-10