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I was trying to solve the following limit,

$$\lim_{x \to 1^+} \frac{x}{\ln(x)}$$

using L'Hopitals rule, but when applying the limits to the given equation, the function turns to the form $\tfrac{1}{0}$ which does not satisfy the required indeterminate form to use L'Hopitals rule. Is my analysis correct?.

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    $\begingroup$ Why do you wnat to use L'Hopital rule? the limit has not and indeterminate form. $\endgroup$ – Emilio Novati Nov 10 '15 at 20:52
  • $\begingroup$ It is. Also, the limit does not exist. $\endgroup$ – B. Pasternak Nov 10 '15 at 20:53
  • $\begingroup$ The expression increases (toward positive infinity) as $x$ approaches $1$ from above, and decreases (toward negative infinity) as $x$ approaches $1$ from below. $\endgroup$ – Brian Tung Nov 10 '15 at 21:15
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As you already found out, the fraction has a finite non-zero numerator (converging to one) and the denominator converging tho zero, hence the limit $$\lim_{x\to 1}\frac{x}{\ln x}$$ does not exist. And, of course, we can not apply l'Hôpital's rule, because it works only for indetermined forms $\frac 00$ or $\frac\infty\infty$.

However, one can say that $$\lim_{x\to 1+}\frac{x}{\ln x}=+\infty\\\lim_{x\to 1-}\frac{x}{\ln x}=-\infty,$$ because logarithm changes its sign at $x=1$.

Another thing you might want to consider is to study the limit $$\lim_{x\to 0}\frac{x}{\ln (1+x)}.$$In this case l'Hôpital's rule works well.

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Remember that you can only use L'Hospital's Rule when you have inderterminations of the form $$ \frac{\infty}{\infty} $$ And $$ \frac{0}{0} $$ If you don't have an indetermination of this form, you can try to re-write your expression such that it has one of the indeterminations above.

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You are correct, you cannot use L'Hopital's rule in this case. The limit from the right is simply $+\infty$.

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