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13. Suppose the function f : R _ R has limit L at 0, and let a > 0. If g : R _ R is defined by g(x) = f(ax) for x € R, show that lim g(x) = L X-0...

Question

13. Suppose the function f : R _ R has limit L at 0, and let a > 0. If g : R _ R is defined by g(x) = f(ax) for x € R, show that lim g(x) = L X-0

13. Suppose the function f : R _ R has limit L at 0, and let a > 0. If g : R _ R is defined by g(x) = f(ax) for x € R, show that lim g(x) = L X-0



Answers

Suppose that $f$ and $g$ are continuous functions such that $f(2)=1$ and $\lim _{x \rightarrow 2}[f(x)+4 g(x)]=13 .$ Find (a) $g(2)$ (b) $\lim _{x \rightarrow 2} g(x)$.

Let's go ahead and consider. We have a function F X, which is greater than or equal to zero. We have that the limit as X approaches A of f of X is equal to zero and the limit as X approaches. A of ah function G of X is equal to negative infinity and when we want to show So wtsp we want to show that the limit as X approaches a off ffx to the G of X is equal to infinity. So let's go ahead and assume that our limit exists and equal toe wise we have y is equal to the limit as X approaches a off affects to the G of X. Now we take the natural log off both sides. So we have the natural log off. Why is equal to while the natural log off off? Well, the limit as X approaches a off ffx to the G of x. Okay, so then we can go ahead. You can take this limit outside off the log, right? So we have the natural log of y is then equal to the limit as X approaches a off of the natural log off after backs to the G of X to the G of X. Okay, So, um then well, we get here. This is equal to the limit as X approaches a off well off of g of X. Right, Because that's a lot to a power. That power should come out front of the log. So you have the limit as X approaches a off G of X, um times the natural log of f of X. Okay, so then well, this is equal to the limit as X approaches A of G of x times the limit as ex approaches a of the natural log of f of acts. Okay, but both of these are equal to negative infinity. Right? This is This is negative. Infinity. This is negative. Infinity times. What's that? Doesn't look like infinity. This is so negative. Yeah, right. That limit is negative. Infinity, This limit is also native infinity. You have native infinity times Negative infinity. Well, that's just infinity. So, um therefore right, The natural log of y natural log of y is equal to infinity. Okay, Um but the national of why, right? So I mean, why all we could I mean, we could say If if this is true, right? That's what that's a lot of Why is logged based E. This means that that why is equal to e to the infinity right, which is equal to one So which is equal to infinity. And why was our limit, right? We let why be equal to the limit as X approaches a of a fax to the G of X. So therefore, that limit is therefore equal to infinity. All right, take care.

Okay. What we're gonna do is walk through a proof all based off of using the epsilon delta definition of a limit. And so we're told if ever vax if the absolute value of F of X um is less than B for for an absolute value of X. My essay is less than one. And the limit as X approaches A of G of X equals zero, then the limit as X approaches A of F of X times G of X equals zero. Um Okay, in a couple of things is we notice um that this actually um is epsilon and this of course would be F of x minus zero for x minus, see less than delta. Right? So this is this right? Here would be the limit as X approaches A of F of X equals zero as well. Right? Um And then of course this right here we're gonna be working with the limit as X approaches A of G of X equal to zero. And so we're going to use that um definition epsilon delta definition. Um So we're going to choose an epsilon greater than zero. Um Since to limit as X approaches a G fx equals zero, then there is a delta one greater than zero such that zero less than the absolute drive, X minus a is less than they sell to one. Which implies that the absolute value of G of x minus zero is less than And we're going to let it be less than some epsilon over be valued. Okay? And we're going to let this delta, since we're working with two different um functions and limits, we're gonna let delta itself equal the minimum of one. Or this delta one value. Okay? Um And so then sorry um then we know that as the vax absolutely F F of X is less than this. Be value for the absolute value of X minus a less than the delta, Right? So if we let one delta be a minimum, and since then the absolute value of F. Of X. S and B for that value, which of course implies then that right? And thus therefore, so now if I multiply um this guy by G of X, we have oops times chair backs and of course both of them are minus zero. Keep that in mind. Um is less than which implies that this ffx times cheer vax is lesson epsilon, which means you look at both of those, then that means that the limit as X approaches A of Activex times, she vax has equal zero.


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