# How to make this limit question a indeterminate form? (L-Hopital)

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

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?

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

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

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

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