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# Given that$$\displaystyle \lim_{x\to a} f(x) = 0$$ $$\displaystyle \lim_{x\to a} g(x) = 0$$ $$\displaystyle \lim_{x\to a} h(x) = 1$$$$\displaystyle \lim_{x\to a} p(x) = \infty$$ $$\displaystyle \lim_{x\to a} q(x) = \infty$$which of the following limits are indeterminate forms? For those that are not an indeterminate form, evaluate the limit where possible(a) $\displaystyle \lim_{x\to a} [f(x) - p(x)]$(b) $\displaystyle \lim_{x\to a} [p(x) - q(x)]$(c) $\displaystyle \lim_{x\to a} [p(x) + q(x)]$

## (a) When $x$ is near $a, f(x)$ is near 0 and $p(x)$ is large, so $f(x)-p(x)$ is large negative. Thus, $\lim _{x \rightarrow a}[f(x)-p(x)]=-\infty$(b) $\lim _{x \rightarrow a}[p(x)-q(x)]$ is an indeterminate form of type $\infty-\infty$(c) When $x$ is near $a, p(x)$ and $q(x)$ are both large, $\operatorname{sop}(x)+q(x)$ is large. Thus, $\lim _{x \rightarrow a}[p(x)+q(x)]=\infty$

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Differentiation

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

So here's what we have is um F of x minus Pierre X. So fx minus P of X is going to give us zero minus infinity and that's just gonna be a negative in family. Don't mind that. It says undefined. That's just because um this calculating tool doesn't recognize infinity as a number but the limit would actually still exist. Then what we have is um P of x minus Q of its subcommittee, infinity minus infinity. This would be considered an indeterminate form. Well, it might seem like you could have that would just be zero. That would not be the case. This would be an undefined value or it would be an indeterminate form then. Lastly we have P of X plus two of X. That would be infinity plus infinity since those are both infinitely large. Um We know that this would just be infinity as well. So you can't think of infinity in terms of a number you can't say, oh, infinity, medicine finished zero. Because anything minus itself, zero infinity takes on different sizes. It's it's not something that we can contain. So it's not something that we can just subtract directly and get zero. It would still be considered an indeterminate form. So when doing low petals role, we would consider this something that we would have to to look at and consider indeterminate forms that way we can use the hotels.

California Baptist University

#### Topics

Derivatives

Differentiation

Volume

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp