`lim_(x->1^+) [ln(x^7 - 1) - ln(x^5 - 1)]` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule...

`lim_(x->1^+) [ln(x^7 - 1) - ln(x^5 - 1)]` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Chapter 4, 4.4 - Problem 54 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the limit but first you need to use the logaritmic property to convert the difference of logarithms into the logarithm of quotient:

`lim_(x->1^+) (ln(x^7-1) - ln(x^5-1)) = lim_(x->1^+) ln((x^7-1)/(x^5-1))`

`lim_(x->1^+) ln((x^7-1)/(x^5-1)) = ln lim_(x->1^+)((x^7-1)/(x^5-1)) = ln 0/0`

Since the limit is indetrminate `0/0` , you may use l'Hospital's rule:

`ln lim_(x->1^+)((x^7-1)/(x^5-1)) = ln lim_(x->1^+)(7x^6)/(5x^4) = ln (7/5)`

Hence, evaluating the limit of the function using l'Hospital's rule yields `lim_(x->1^+) (ln(x^7-1) - ln(x^5-1)) = ln (7/5).`

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