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I want to find the limit of this example using L'Hospital rule i get easily ans. but i want to find the limit without using L'Hospital $$ \lim_{x\to 1}\frac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e} $$ I tried to set the power of for using some formula of limit but after the what can i do with $e^x-e$ ?

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    Why do you want to avoid l'Hôpital here? The problem seems to be constructed explicitly to make using l'Hôpital's rule about a hundred times easier than anything else (assuming there is any way to evaluate it at all that isn't just l'Hôpital in disguise).2012-06-26
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    I'd probably put $h=x-1$ and use some expansion in Taylor series, but this would be less elementary and almost equivalent to the use of De l'Hospital's rule.2012-06-26
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    Can everyone please stop putting an s in poor Mr. de l'Hôpital's name? He is not a big building with patients :(2012-06-26
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    Sorry, but I believe that the name with the *s* is acceptable.Actually, this is how the "marquis" used to write his own name ;-) Moreover, the english word "hospital" is the french word "hôpital", I guess.2012-06-26
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    Wikipedia says "In the 17th and 18th centuries, the name was commonly spelled "l'Hospital", however, French spellings have been altered: the silent 's' has been dropped and replaced with the circumflex over the preceding vowel. The former spelling is still used in English where there is no circumflex." Today I learnt something! (In fairness, I think the s is still silent, so it's probably less misleading to leave it out).2012-06-26
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    I saw a signature by this guy, it was more like *Lhospital* with no apostrophe!2012-06-26

4 Answers 4

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Supposing you don't know what a derivative is...

Writing $$ \frac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e} = \frac{\sqrt[n] x + 2} {e n} \frac {e^{\frac{\ln x} n} - 1} {\frac {\ln x} n} \frac {\ln x} {x - 1} \frac {x - 1} {e^{x - 1} - 1} $$

the above limit is reduced to a product of "special" limits.

Edit A more direct way is to write $$ \frac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e} = \frac{\sqrt[n] x + 2} {e} \frac {\sqrt[n] x - 1} {x - 1} \frac {x - 1} {e^{x - 1} - 1} $$ Since $u^n - 1 = (u - 1) (u^{n - 1} + \dotsb + 1)$, setting $u = \sqrt[n] x$, we have $$ \frac {\sqrt[n] x - 1} {x - 1} = \frac {u - 1} {u^n - 1} = \frac 1 {u^{n - 1} + \dotsb + 1} \to \frac 1 n $$

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    This seems overly complicated. Why not simply $\dfrac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e}=\dfrac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{x-1}\dfrac{x-1}{e^x-e}$ ?2012-06-26
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    @lhf: Because to calculate the values of the two limits on the right you have to know function derivatives and I want to give an answer without reference to that notion.2012-06-26
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    How does one calculate the limits of $\displaystyle \frac{\ln x}{x-1},\frac{x-1}{e^{x-1}-1}$ without knowing what a derivative is?2012-06-26
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    @anon: Setting $y = 1/x$, $\displaystyle \lim_{x\to 0}\frac {\ln (x + 1)} x = \lim_{y\to \infty} \ln \left(1 + \frac 1 y \right)^y = \ln e = 1$2012-06-26
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    @anon: If you know derivative and its rules it's trivial to calculate in one step the limit of $\dfrac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{‌​x-1}$. If you don't know them, you have to decompose it as I did.2012-06-26
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    Ah, of course.${}$2012-06-26
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Since $n$ is fixed, we can say $x\to1\iff u:=x^{1/n}\to 1$, and hence substitute

$$\lim_{x\to 1}\frac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e}=\frac{1}{e}\lim_{u\to 1}\frac{u^2+u-2}{e^{\large u^n-1}-1}$$

We factored an $e$ out the denominator. Further, we can factor the numerator as $(u-1)(u+2)$, and of course $u+2\to3$: pull this out of the limit and the resulting limit expression will be the reciprocal of a derivative of a certain function at $u=1$...

(Seriously though, what's wrong with good ol' l'Hospital's rule?)

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You can rewrite the limit as $$\lim_{x \rightarrow 1} \bigg({x^{1 \over n} + x^{2 \over n} - 2 \over x - 1} \bigg/ {e^x - e \over x - 1}\bigg)$$ $$= \bigg(\lim_{x \rightarrow 1} {x^{1 \over n} + x^{2 \over n} - 2 \over x - 1} \bigg/ \lim_{x \rightarrow 1}{e^x - e \over x - 1}\bigg)$$ $= {\displaystyle {f'(1) \over g'(1)}}$, where $f(x) = x^{1 \over n} + x^{2 \over n}$, and $g(x) = e^x$, using the definition of derivative.

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    L'Hopital in disguise! =)2012-06-26
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    Well... this time it really is just the definition of derivative in terms of difference quotients.2012-06-26
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    @Zarrax: Just because you provide a proof of the rule does not prevent the rule you prove (a special case of) from _being_ l'Hospital's rule.2012-06-26
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    @HenningMakholm He asked to not use L'Hopital's rule. I didn't. Since the proof of l'Hopital's rule in this situation seemed to be the easiest way to find the limit I decided to write it out. I've seen this phenomenon a few times on this site now, where it's actually much easier to use the proof of l'Hopital than avoid it... so I like to emphasize this.2012-06-26
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    Whatever. The OP asked for a solution in which the rule is not being applied. You provided an answer in which the rule _is_ applied. The fact that you also prove it does not mean that the subsequent application ceases to be an application of the rule.2012-06-26
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    @HenningMakholm His words were "i want to find the limit without using L'hospital " I'm not really interested in debating the exact meaning of this. My main issue in this problem (and similar ones) is that this kind of argument is really a good way to solve it, and I don't think one should avoid using such methods.2012-06-26
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$$ \begin{aligned} \lim_{x\to 1}\left(\frac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e}\right) & = \lim _{t\to 0}\left(\frac{\sqrt[n]{t+1}+\sqrt[n]{\left(t+1\right)^2}-2}{e^{t+1}-e}\right) \\& = \lim _{t\to 0}\left(\frac{\left(1+\frac{t}{n}+o\left(t\right)\right)+\left(1+\frac{2t}{n}+o\left(t\right)\right)-2}{\left(e+et+o\left(t\right)\right)-e}\right) \\& = \color{red}{\frac{3}{en}} \end{aligned} $$ Solved with substitution $t = x-1$ and Taylor expansion