On evaluation of $\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$ Evaluate the following limit:$$\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$$

Answer is $\frac{1}{\sqrt{e}}$ as on Wolfram Alpha query and evaluated by sympy (Python). Also here is the math search engine query which returned nothing substantial.

I think the answer is $1$. Why?
If $f(x)\to1$ and $g(x)\to\infty$ then the limit ${f(x)}^{g(x)}=e^{({f(x)-1})\cdot g(x)}$. For those who are curious of the proof:

 $$\lim_{x\to a} [1+(f(x)-1)]^{\frac{1}{f(x)-1} (f(x)-1)g(x)} = \lim_{x\to a} [(1+(f(x)-1))^{\frac{1}{f(x)-1}}]^{\lim_{x\to a} (f(x)-1)g(x)}$$


and then a very popular limit to sum it up $\lim_{x\to0} (1+x)^{\frac{1}{x}}=e$

Applying the limit on the numerator, we get $e^{\frac{1}{h}}$ which is exactly equal to the denominator. So hence, the answer $1$. Only my close friend desmos supports me here:

Whereas Wolfram shows


$$\text{My Questions}$$
$1$. Why does the method I adopted shows a speculatively wrong result?
$2$. Is there any way to correct it?

Source code for sympy evaluation.
#Python 3.8
import math
from sympy import * 
  
x = symbols('x') 
expr = ((1+x)**(1/x**2))/(math.e)**(1/x); 
    
print("Expression : {}".format(expr))  
      
# Use sympy.limit() method  
limit_expr = limit(expr, x, 0)   
      
print("Limit of the expression tends to 0 : {}".format(limit_expr)) 

Run the code:
Expression : 2.71828182845905**(-1/x)*(x + 1)**(x**(-2))
Limit of the expression tends to 0 : 0.606530659712633

 A: For your first question see Daniel's comment. To get W|A's answer we just need L'Hopital:$$\lim_{h\to0}\left(\frac{1+h}{e^h}\right)^{1/h^2}=\exp\lim_{h\to0}\frac{\log((1+h)/e^h)}{h^2}=\exp\lim_{h\to0}\frac{\frac{e^h}{1+h}\cdot\frac{-h}{e^h}}{2h}=\frac1{\sqrt e}.$$
A: In this case we have
$$\left(\frac{{1+h}}{e^h}\right)^\frac1{h^2}=\left[\left(1+\left(\frac{{1+h}}{e^h}-1\right)\right)^\frac{1}{\frac{{1+h}}{e^h}-1}\right]^{\frac{\frac{{1+h}}{e^h}-1}{h^2}}$$
with
$$\left(1+\left(\frac{{1+h}}{e^h}-1\right)\right)^\frac{1}{\frac{{1+h}}{e^h}-1} \to e$$
and
$$\frac{\frac{{1+h}}{e^h}-1}{h^2}=\frac1{e^h}\frac{1+h-e^h}{h^2} \to -\frac12$$
A: $$A= \sqrt[h^2]\frac{{1+h}}{e^h}\implies \log(A)=\frac 1{h^2}\left( \log(1+h)-h\right)$$ Now, by Taylor
$$\log(A)=\frac 1{h^2}\left(-\frac{h^2}{2}+\frac{h^3}{3}+O\left(h^4\right) \right)=-\frac{1}{2}+\frac{h}{3}+O\left(h^2\right)$$
$$A=e^{\log(A)}=\frac{1}{\sqrt{e}}+\frac{h}{3 \sqrt{e}}+O\left(h^2\right)$$
A: I assume that you are familiar with L'Hospital's Rule for evaluating limits.
Applying the formula which you mentioned in the question,
$$lim_{h\to0}\frac{e^{-h}(1+h) - 1}{h^2} = \frac{-1}{2}$$
by the application of L'Hospital's Rule.
So the final limit is $\frac{1}{\sqrt(e)}$
