so I have an issue and I have to solve this limit $$\lim_{x\to -\infty}x^x$$ so I worked on it and tried to solve it.
My attempt:
Let there be a second variable $y=-x$.So we have this $$\lim_{x \to -\infty}y=\lim_{x \to -\infty}-x=+\infty$$ So when $$ x \to -\infty \Rightarrow y \to +\infty$$Now working with the actual limit$$\lim_{x \to -\infty}x^x=\lim_{x \to -\infty}\frac{-1}{-x^{-x}}$$ but $$-x=y$$ so $$\lim_{x \to -\infty}x^x=\lim_{y \to +\infty}\frac{-1}{y^y}=0$$ because $$\lim_{x\to +\infty}x^x=+\infty$$
Can I do all these or not?? Thanks for your time
The short version is you've made a grouping error. It's correct to say $$\lim_{x\to-\infty}x^x = \lim_{x\to-\infty}\frac{-1}{-(x^{-x})}$$ but you have to remember that the minus sign in the denominator is outside the exponent. You implicitly pull it inside when you switch in $y$, and that's where you went wrong. A substitution you can make, however, is $$\lim_{x\to-\infty}x^x = \lim_{y\to+\infty}(-y)^{-y} = \lim_{y\to+\infty}e^{-y\ln(-y)}$$ The natural log of a negative number is a multivalued function, but in this case, you'll see that any choice gives the same answer, so for now we'll write $$\lim_{y\to+\infty}e^{-y(\ln y + k\pi i)}$$ where $k$ is an arbitrary odd integer. (I'm assuming you're familiar with the identity $e^{ix} = \cos x + i\sin x$; if not, well, there it is.) So you end up with $$\lim_{y\to+\infty}e^{-y\ln y }e^{-k\pi iy} = \lim_{y\to+\infty}\frac{e^{-k\pi iy}}{y^y}$$ The denominator obviously blows up, whereas the numerator stays on the unit circle, so the limit is zero.