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Uh el mel lim UKOH I...

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Uh el mel lim UKOH I

Uh el mel lim UKOH I



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$\lim _{(x, y) \rightarrow(0,0)} e^{x^{2}+y^{2}}=$ __________.

Are there. So we're asked again about limits and we see here that in this section we've been working it quite a bit with, uh, with chicken medical functions as well as, uh oh, no meals and the ah, the trick here essentially developed this limits, he goes zero of sine squared three t over T squares is to employ the direct substitution property, which we said it applied for both to conduct reconnaissance and polynomial is rational functions. Um, morning exercises in the section is indeed to prove that the drug substation property applies for trademark it functions. Um, but this will also involve another trick, namely, ah, identifying the limit of this product. You see the square here as a product of limits. So what we do is we break it into two. So we have the limit. As she goes to zero of signed three t over tea squared, the whole thing squared. So in one step, what I've done is I've identified that we can rewrite this as sign of three t over t squared and then I brought the limits. I brought the limit into the product. Now this is almost in a former recognized because you've also done exercises where we have something of a form. FluMist. He goes to zero of a sign of three T over three teeth. So let's find a way to manipulate this. So we have that same form, and then we'll be able to identify that when it's in that form, we have something that equals one. I believe that was expressed an equation six in this section. Um so let's observe that if we write, um, limit as he goes to zero of signed three tea over three t and we write that whole thing squared limit SC goes to zero of sign of three t perhaps sign of three t over three T than in order for this expression here to equal the one above, we need to add a factor of nine and front so that we get something that cancels, and I'm slightly employing the fact that we can take the constant out of a limit without changing it. So, for instance, you could take these threes out, and then you have nine over nine. Um, but any rate, then we see that we have seen that looks like nine times and analysis in this form that we recognize one. So we have nine times one times along. So we have nine, so our limit is not him.

So we have to evaluate the following limit and see if it equals one over E. So um the first thing to do is to plug in the value of the limit and see if it's something easy. So if I plug in the value H equals zero. I get one minus zero raised to the negative 1/0. And since you can't divide by zero, this becomes a problem. Now in order to solve this problem, we need to use one rule called low petals rules. So I just want to remind everyone what low petals rule is. So let's suppose that you have a function actually a quotation of two functions and you're taking the limit as X approaches some about you And you can write the expression as a ratio of two functions. Now if this evaluates to an indeterminate form that either looks like 0/0 or infinity over infinity and this is plus or minus infinity. Then you can rewrite the limit as follows. You can take the limit of the derivative of the numerator divided by the derivative of the denominator. So I just want to quickly interject and say that this is not the corset rule. Look up the quotient rule, remind yourself that is nothing that has nothing to do with this. This just says that you isolate F and take the derivative of that and then you take the derivative of G. And then you compare the slopes. Okay? So it essentially gives you an easier way of evaluating the limit because usually the derivative is a slightly simpler function, especially with polynomial. So that being said, let's evaluate this limit. The first thing I'm gonna do is I'm actually going to set the limit equal to evaluate why. You'll see why in just a second. So the limit h approaches zero of one minus H. To the negative one over H. I'm going to let that equal why. And the reason is this exponent is quite pesky in order to get rid of the expo and I'm actually gonna take the natural log of both sides. Now you can interchange a function and a limit as long as the function is continuous and in this case the natural log is continuous. So I'm gonna go ahead and say that the natural log of Y equals the limit as a jew approaches zero of the natural log of one minus H. To the negative one over H. Now, because the properties of logarithms, I can move this exponent down as a coefficient and then I get that the natural log of y is equal to the limit. As a jew approaches zero of negative one over H times the natural log one minus H. Just to remind you, we are trying to solve for why why is equal to the limit that we want? This is why. And here's the limit that we want in blue. Now to evaluate this limit. Um I can rewrite this as the limit As a church approaches zero of the Natural Log. Well negative the natural log of 1 -3. All over H. Okay. And if I plug in H equals zero at this point so I go and just plug that in, I will get negative natural log of 1 0 over zero. Well the natural log of 1 0 is the natural log of one And the natural log of one is 0. So I was able to write a ratio of two functions. And upon evaluating the limit, I get an indeterminate form. So now I can use low P. Tall and lo petals rule says that the same limit which was equal to the natural log of Y. Is the same limit. But now I separately take the derivative of the top and the bottom. So I'm actually just going to write that out. Just we don't confuse it with the caution rule. And I should specify that I actually made a mistake. Um It's not a big mistake here. We're taking the derivative with respect to H. I never use H. That's kind of line thrown up. I always say as X approaches something. So regardless, anytime you take a derivative, uh this variable in this variable better match. So that was my bad. Okay so this is just gonna be H. Now the derivative with respect to H. Of negative natural log of whatever. I can factor the negative out. And the derivative of the natural log is just one over whatever's inside. But I have to remember to multiply by the inside function using the chain rule. So I also have to take the derivative of um one minus H. And that's just negative one. So these negatives cancel and the derivative of um H. Is just one. So I have to remember that I can take the same limit but now I evaluated those derivatives. Okay, so just to rewrite what we've done the natural log of why we use loopholes rule in the numerator and denominator. And we said that now I'm going to take the limit, I'll be slightly simpler function. So these negative signs go away and I'm left with 1/1 -3 divided by one. Well that limit is super easy because now I can directly substitute the value at H equals zero And I get 1/1 0 which is one. I'm not done. The limit does not equal one. Because the limit that I'm interested in is why I set the limit equal to Y in the very beginning. So the natural log of Y is equal to one. So if I exponentially both sides, I get that. Either the natural log of Y cancels to why and the limit that I'm interested in why is equal to E. To the first power, which is just so this is the limit and it's actually not equal to um one over Eats. This is false.

Okay. What we wanted Thio is to determine the limit as why approaches one, uh, seek int. Why? Time seeking squared. Why? Minus tangent squared? Why? Minus one? And the first thing you should do when you're trying to evaluate limits is to do the direct substitution. And so that's what we're gonna do. And so we have seeking of, um, seek it squared of one minus tangent, squared of one minus one. Now we're gonna go ahead and you need to go ahead. And what we do is put all of this in our calculator. One thing you need to remember is your calculator needs to be in radiant note. And most calculators do no, have, um, seek it in the calculator. And so you're gonna have to put seeking in as one over co sign. And when you do that on your calculator, you get this equal to one. And so since I get a value for that direct substitution, that means the function is continuous at why equal toe? Why

If we were toe substitute t a zero, we'd get 0/0. And so we'll rearrange. That's by rewriting tangent. And that's the same thing is sign of tea overcoat sine of T, which would then flip back up to the numerator. And then, of course, we still have the to t up top. And so now we could then move this to as a constant in front and then recognizing that t divided by sine of T as t approaches. Zero is old one we could direct substitute zero for tea and co sign of zero would also give us one. And so, in the end, we have two times are ones there or a final solution of two.


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