$\lim_{x\to\infty}(\frac{x^2}{x^2+x})^x = 1 $ or $0$? $$\lim_{x\to\infty}\left(\frac{x^2}{x^2+x}\right)^x$$
as $\lim_{x\to\infty}\frac{x^2}{x^2+x} = 1$ but we all know that $\frac{x^2}{x^2+x} < 1$ and for $a<1$ lim $\lim_{x\to\infty} a^x = 0$
I don't know how to find the limit in this case

Mathlove privided an efficient way forward.  Here is a slightly different approach.  Take the logarithm of the function of interest.  Then, we have 
$$\log\left(\frac{x^2}{x+x^2}\right)^x=x\log\left(\frac{x}{x+1}\right)=\frac{\log\left(\frac{x}{1+x}\right)}{\frac{1}{x}}=\frac{\log x- \log (1+x)}{\frac{1}{x}}$$
Now, apply L'Hospital's Rule to find
$$\lim_{x\to \infty} \log\left(\frac{x^2}{x+x^2}\right)^x=\lim_{x\to \infty} \frac{\frac{1}{x}- \frac{1}{1+x}}{-\frac{1}{x^2}}=-1$$
Since the exponential function is continuous, then we have 
$$\lim_{x\to \infty} \left(\frac{x^2}{x+x^2}\right)^x=e^{-1}$$
