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$$\lim_{x\to 1^+}[\ln(x^7 -1) - \ln(x^5 -1)]$$

This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $\ln(\frac{0}{0})$ might not be right way to convert it into indeterminate form.

How to solve this question?

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$$\lim_{x \to 1^+} [\ln(x^7-1)- \ln(x^5-1)]=\ln \left[\lim_{x\to 1^+} \frac{x^7-1}{x^5-1} \right]$$

Evaluate $\lim_{x\to 1^+} \frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.

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  • $\begingroup$ I am getting the answer ln(7/5). Is it right? $\endgroup$ – Amogh Joshi Nov 14 '18 at 6:49
  • $\begingroup$ yes, congratulations. $\endgroup$ – Siong Thye Goh Nov 14 '18 at 6:51
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Hint:

$$\lim_{x\to 1^+}\left[\ln(x^7 -1) - \ln(x^5 -1)\right]=\ln\left(\lim_{x\to 1^+}\dfrac{x^7-1}{x^5-1}\right)$$

Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$

Alternatively, $\dfrac{x^7-1}{x^5-1}=\dfrac{\dfrac{x^7-1}{x-1}}{\dfrac{x^5-1}{x-1}}$

and $$\lim_{x\to 1}\dfrac{x^n-1}{x-1}=\dfrac{d(x^n)}{dx}_{\text{( at }x=1)}$$

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