# What is the limit without using L'hospital?

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 $e^x-e$ ?

-
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). – Henning Makholm Jun 26 '12 at 11:36
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. – Siminore Jun 26 '12 at 11:37
Can everyone please stop putting an s in poor Mr. de l'Hôpital's name? He is not a big building with patients :( – Ben Millwood Jun 26 '12 at 12:15
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. – Siminore Jun 26 '12 at 12:20
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). – Ben Millwood Jun 26 '12 at 13:00

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 $$\frac {\sqrt[n] x - 1} {x - 1} = \frac {u - 1} {u^n - 1} = \frac 1 {u^{n - 1} + \dotsb + 1} \to \frac 1 n$$

-
This seems overly complicated. Why not simply $\dfrac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e}=\dfrac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{‌​x-1}\dfrac{x-1}{e^x-e}$ ? – lhf Jun 26 '12 at 12:50
@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. – AlbertH Jun 26 '12 at 14:48
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? – anon Jun 26 '12 at 15:36
@anon: Setting $y = 1/x$, $\displaystyle \lim_{x\to 0}\frac {\ln (x + 1)} x = \lim_{y\to \infty} \ln \left(1 + \frac 1 y \right)^y = \ln e = 1$ – AlbertH Jun 26 '12 at 16:01
@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. – AlbertH Jun 26 '12 at 16:15

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?)

-

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.

-
L'Hopital in disguise! =) – Pedro Tamaroff Jun 26 '12 at 14:20
Well... this time it really is just the definition of derivative in terms of difference quotients. – Zarrax Jun 26 '12 at 15:17
@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. – Henning Makholm Jun 26 '12 at 16:32
@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. – Zarrax Jun 26 '12 at 16:45
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. – Henning Makholm Jun 26 '12 at 17:11