So I have this limit of $x$ tending to $0^+$. Of: $$ e^{(-1/x)}\ln(1/x^2) $$ I know that I can solve this by using subsitution, thus substituting $1/x$ with $u$, but I really don't seem to grasp the concept, and thus the method in order to solve the system.
I also know that I can solve it using L'Hospital's rule, even if it may seem more difficult, but could I do this: could I convert the function into a fraction and then take the derivative of the numerator and the derivative of the denominator? Could you please show me some steps? Thanks!
I would be really, really useful if you could show me both types of ways on how to solve this limit. (If I had to choose, I would prefer some steps regarding the use of L'Hospital's rule)
Sure and thanks to the laws of exponents, it is easily turned into a fraction: $$e^{-\frac{1}{x}}\ln\frac{1}{x^2}=\frac{\ln\frac{1}{x^2}}{e^{\frac{1}{x}}}$$
Both numerator and denominator now tend to $\infty$ for $x \to 0^+$ so you can apply l'Hôpital's rule: $$\lim_{x \to 0^+}\frac{\ln\frac{1}{x^2}}{e^{\frac{1}{x}}}=\lim_{x \to 0^+}\frac{\left(\ln\frac{1}{x^2}\right)'}{\left(e^{\frac{1}{x}}\right)'}=\lim_{x \to 0^+}\frac{\frac{-2}{x}}{-x^{-2}e^{\frac{1}{x}}}=\lim_{x \to 0^+} \left( 2\,x\,e^{-\frac{1}{x}}\right)=\ldots$$