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Numerade Educator

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Problem 4 Medium Difficulty

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)]^{g(x)} $
(b) $ \displaystyle \lim_{x\to a} [f(x)]^{p(x)} $
(c) $ \displaystyle \lim_{x\to a} [h(x)]^{p(x)} $
(d) $ \displaystyle \lim_{x\to a} [p(x)]^{f(x)} $
(e) $ \displaystyle \lim_{x\to a} [p(x)]^{q(x)} $
(f) $ \displaystyle \lim_{x\to a} \sqrt[q(x)]{p(x)} $

Answer

a) indeterminate
b)$\lim _{x \rightarrow a}[f(x)]^{p(x)}=0$
c) indeterminate
d) indeterminate
e) $\lim _{x \rightarrow a} p(x)^{\phi(x)}=\infty$
f) indeterminate

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Video Transcript

Hello. So here, in part A we consider the following limit where we have the limit as X approaches A of F of X to the G M X. Well, this is going to be equal. This is of the form here. Um zero to the zero, which we know is an indeterminate form. So therefore the given limit here. The limit as X approaches A of F of X to the G Fx is an indeterminate form. So yes, this is an indeterminate forum for part B. We consider the limit as those are the limit. As X approaches A of F of X to the P of X, we have alfa backs to the P of X. Um So let's go ahead and let y equal F of X to the p M X. And then apply the law algorithm to both sides. So therefore we have the natural log of Y is going to be equal to. Well, the natural log of F of X to the P M X is equal to P M X times the natural log of affects right log to power. Power comes outfront as a coefficient we have your backs times the natural log of fx. Okay. And then we can then apply the exponential on both sides. E to the natural log of Y is just equal to Y. To get Y is equal to well E to the P of X times the natural log of fx. Um and then we get while F of X to the P of X is equal to E to the P of x times the natural log of fx. Um So again, applying the log on both sides gives us, we take the limit as X approaches A of F of X to the PM backs that is equal to the limit as X approaches A. Of E to the P of x times the natural log Activex. Um which is another form here, um E to the infinity minus infinity. Um which in the limit that's going to be equal to E to the negative infinity, um which is going to be equal to zero. So therefore this given limit is not is not um an indeterminate form and we evaluated here and we get zero. Um If apart, see, we consider the limit as X approaches A of H of X to the P. Of X. So then we have while the limited exports is A of H of X uh to the P of X. Um This is going to be equal to one to the infinity um which is an indeterminate form. So yes, yes, this would be an indeterminate form. Um par de we consider the limit um as exports A of P of X to the F. Of X. So now we have the limit as X approaches A of P of X to the F. Of X. Um So here again, we get to form infinity to the zero, which is an indeterminate form. This is gonna be another form infinity to the zero. And yes, infinity to zero is an indeterminate form. So, yes. And part E. We have the limit as X approaches A of P of X to the Q. Of X. So for part E we have the limit as X approaches A of P. Of X to the Q. Of X. Um And this is just gonna be equal to infinity. So therefore this is not not an indeterminate form. And lastly, we have the Q fruit of p vivax. We take the limit part F. The limit as X approaches A. Of um the well the show with the Q of X route of P of X. Um Okay, so this is going to be equal to the limit as X approaches A. Of what well of to the Q of X. Right? This is the same thing as saying P vivax to the one over Q of X power, um which we get other form infinity to the zero. So this is going to be equal to infinity to the zero power, which is an indeterminate form. So yes, this is and some reform.