2 years ago

I don't know the answer

Mathematics

1 answer:

RUDIKE [14]2 years ago

6 0

Step-by-step explanation:

okay hold up a sec

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Answer:

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1 year ago

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2 years ago

Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i

liberstina [14]

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.

<h3>Limit</h3>

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.

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