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# 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.$\lim_{x\to 1} \frac{x^a - 1}{x^b - 1}$, $b \not= 0$

## This limit has the form$\frac{0}{0} \cdot \lim _{x \rightarrow 1} \frac{x^{a}-1}{x^{b}-1} \quad[\text { for } b \neq 0] \quad \stackrel{\Perp}{=} \lim _{x \rightarrow 1} \frac{a x^{a-1}}{b x^{b-1}}=\frac{a(1)}{b(1)}=\frac{a}{b}$

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

mhm. There's probably want to take the limit of the function. Um and possibly use Law Patel's rule to evaluate it. So in this case we have the limit of the numerator. Uh regardless we're going to get 1 -1 so we will get 0/0, which is going to be and the determinant form. So we want to take the derivative of the numerator and derivative of the denominator. When we take the derivative of the numerator, we get eight times acts To the A -1. And then in the denominator we get B. X To the B -1. Now when we plug in one we get X to the one minus one, X. To the one minus one, that's just going to be to the zero power. So ultimately where we're going to be getting as our answer is A over B. And that will be the limit as X approaches one of our function.

California Baptist University

#### Topics

Derivatives

Differentiation

Volume

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp