limit of the function $x \to 1$

Question

Why $$\lim_{x\rightarrow1} \frac{ e^{ \frac{1}{\pi} \ln x } -1} {\frac{1}{\pi} \ln x }=1$$ I do not get it at all... I do not want explanation that would involve use of l'hospital rule

Answer 1

If you know that $\lim_{x \to 0}\left(1 + x\right)^{1/x} = e$

Put $\ln y = \dfrac{1}{\pi}\ln x$. The limit becomes $$\lim\limits_{y\to 1}\dfrac{y-1}{\ln y}$$

$z=y-1$ $$\lim\limits_{z\to 0}\dfrac{z}{\ln (z+1)}$$

Also you can obtain $$\lim_{z \to 0}\frac{\ln(1 + z)}{z} = \lim_{z \to 0}\frac{1}{z}\ln(1 + z) = \lim_{z \to 0}\ln((1 + z)^\frac{1}{z}) = \ln e = 1$$

And that means your limit is also equal to $1$.

Answer 2

Hint:

Take $z = \frac{1}{\pi}\text{log}(x)$ and use the definition of derivative.

Answer 3

Let $f$ the function $f(x)=\dfrac{1}{\pi}\ln x$. You have then $f(1)=0$

$\lim\limits_{x\to 1}\dfrac{e^{\frac{1}{\pi}\ln x}-1}{\dfrac{1}{\pi}\ln x}=\lim\limits_{x\to 1}\dfrac{e^{f(x)}-e^{f(1)}}{f(x)-f(1)}$

Answer 4

Without using derivative:

Put $X=\dfrac{1}{\pi}\ln x$. When $x\to 1$, $X\to 0$. The limit becomes:

$\lim\limits_{X\to 0}\dfrac{e^X-1}{X}=1$.