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I'm having trouble simplifying the result of complicated limits which contain mutliple fractions. I understand this is basic math level, but i find it difficult to find exercices to practice that specific problem.

So assuming my start limit goes like this

$$\lim_{x\to1} \frac 1 {ln(x)} - \frac {1} {(x-1)^2}$$

Form : $\infty$ - $\infty$

I put both on the same denominator

$$\lim_{x\to1} \frac {(x-1)^2 - ln (x)} {ln(x)*(x-1)^2} $$

Form : $\frac 0 0$

Hospital rule

$$\lim_{x\to1} \frac {2(x-1)- \frac 1 x} { \frac 1 x*(x-1)^2+ln(x)*2(x-1))}$$

Now i don't know if i can just cancel both $ \frac 1 x$ since there is + and -. Same goes for the $2(x-1)$.

I know that i can transform the $ \frac 1 x * (x-1)^2$ in $\frac {(x-1)^2} {x}$ but this seems like extra steps.

Thanks!

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    $\begingroup$ I would get rid of the compound fraction action first by multiplying top and bottom by $x$. $\endgroup$ – randomgirl Jul 11 '16 at 2:58
  • $\begingroup$ The first form is $\infty - \infty$. $\endgroup$ – Triatticus Jul 11 '16 at 2:58
  • $\begingroup$ Fixed the first form, thanks! For the compound fraction, could you explain a bit more? $\endgroup$ – Maude Jul 11 '16 at 3:08
  • $\begingroup$ I would now inspect what happen if $x\to 1^-$ and $x\to 1^+$. $\endgroup$ – callculus Jul 11 '16 at 3:13
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    $\begingroup$ Note that a tiny bit below $1$, and also a tiny bit above $1$, the after L'Hospital denominator is small positive. And the numerator is close to $-1$. Lesson: simplifying algebraically may not be necessary, looking may do the job. $\endgroup$ – André Nicolas Jul 11 '16 at 3:34

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