Limit of function with taylor series

Question

I am trying very hard to understand the limits with the help of taylor expansion but i still stuck in it ,

$$\lim _{x\to 0^+}\left(e^{\frac{x^2-1}{x}}\right)$$ if I apply the taylor expansion $$1+\frac{x^2-1}{x}+\frac{\left(\frac{x^2-1}{x}\right)^2}{2}+O\left(x^3\right)$$ then applying limit as $x\to 0^+$ will give $\infty$ (infinity) but if i do with limit chain rule i am getting right answer which is $0$.

Answer 1

First a correction: The Taylor expansion (around zero) of the exponential is

$$e^z = 1 + z + \frac{z^2}{2} + O(z^3)$$

Then, in your case you'd get

$$e^{(x^2-1)/x}= 1 + \frac{x^2-1}{x} + \frac{(x^2-1)^2}{2x^2} + O((\frac{x^2-1}{x})^3)$$

But this lead us nowhere. Because we are not interested in $z \to 0^+$, but in $x\to 0^+$ , which implies $z\to -\infty$

Then, it's as simple as recalling $\lim_{z\to -\infty} e^z =0$

Answer 2

What you wrote is not a Taylor series but the start of the infinite series representation of $e^z$.

May be simpler would be $$y=e^{\frac{x^2-1}{x}}\implies \log(y)=\frac{x^2-1}{x}=x-\frac 1x$$ making $$\lim _{x\to 0^+} \log(y)=-\infty\implies\lim _{x\to 0^+}y=0$$