Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using it. lim x→1 [ln(x5 − 1) − ln(x3 − 1)]

Sagot :

Answer:

  ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

Limit

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

  [tex]\diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}[/tex]