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 $e^x-e$ ?
What is $ \lim_{x\to 1}\frac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e} $ without using L'Hospital?
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0I saw a signature by this guy, it was more like *Lhospital* with no apostrophe! – 2012-06-26
4 Answers
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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0Ah, of course.${}$ – 2012-06-26
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.
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0@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
$ \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