What is $ \lim_{x\to 1}\frac{\sqrt[n]{x}+\sqrt[n]{x^2}-2}{e^x-e} $ without using L'Hospital?

Question

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 $e^x-e$ ?

Answer 1

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

Answer 2

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?)

Answer 3

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.