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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)

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$$ e^{-1/x}\ln(1/x^2) $$ 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?

Sure and thanks to the laws of exponents, it is easily turned into a fraction: $$e^{-\frac{1}{x}}\ln\frac{1}{x^2}=\frac{\ln\frac{1}{x^2}}{e^{\frac{1}{x}}}$$

(If I had to choose, I would prefer some steps regarding the use of L'Hospital's rule)

Both numerator and denominator now tend to $\infty$ for $x \to 0^+$ so you can apply l'Hôpital's rule: $$\lim_{x \to 0^+}\frac{\ln\frac{1}{x^2}}{e^{\frac{1}{x}}}=\lim_{x \to 0^+}\frac{\left(\ln\frac{1}{x^2}\right)'}{\left(e^{\frac{1}{x}}\right)'}=\lim_{x \to 0^+}\frac{\frac{-2}{x}}{-x^{-2}e^{\frac{1}{x}}}=\lim_{x \to 0^+} \left( 2\,x\,e^{-\frac{1}{x}}\right)=\ldots$$

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  • $\begingroup$ Thank you SO MUCH . Really . Only one thing i didn't quite grasp. How can e^(-1/x) be re-written as in order to arrive to the first thing you wrote down ? $\endgroup$ – Michael Morello Jun 19 '18 at 14:54
  • $\begingroup$ That's one of the common laws of exponents (you should know/memorize): $a^{-b}=\frac{1}{a^b}$; apply here with $a=e$ as base and exponent $b=\frac{1}{x}$ to get $e^{-1/x}=\frac{1}{e^{1/x}}$. $\endgroup$ – StackTD Jun 19 '18 at 14:56
  • $\begingroup$ Thanks ... last thing, sorry to annoy you , what about the ln ?? originally , would it be just ln/1 ? so that when i multiply out with the e i get what you wrote $\endgroup$ – Michael Morello Jun 19 '18 at 15:00
  • $\begingroup$ $\ Lim_{x → 0^+} e^{-1/x}\ln {1}{x^2}= \ Lim_{x → 0^+} \frac{\ln {1}{x^2}}{e^{1/x}}= 0$ because $\ Lim_{x → 0^+}e^{1/x}= ∞$ $\endgroup$ – sirous Jun 19 '18 at 15:02
  • $\begingroup$ @MichaelMorello I don't understand what you mean with that last question. What is "ln/1"...? $\endgroup$ – StackTD Jun 19 '18 at 15:03

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