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10. Question * (1 Point)lim f} e dx= I30DNESe 4 e [...

Question

10. Question * (1 Point)lim f} e dx= I30DNESe 4 e [

10. Question * (1 Point) lim f} e dx= I30 DNE S e 4 e [



Answers

$$ \lim _{x \rightarrow 0} e^{\operatorname{sgn} x}\left\{\text { Ans. } e, \frac{1}{e}\right\} $$

Okay, so here is equal to one. B is equal to 10 and with end some intervals, we have Delta X is equal to 10 minus one, divided by end, which is nine divided by m. So if we have zero is equal to one. X one is equal to one plus nine over end X two is equal to one plus 18 18 over end and in general, except I is equal to one plus nine. I divided by So therefore we can write f off except eyes equal to one plus nine. I divided by an minus four lm off one plus nine I divided by Since it is not mentioned which point to consider, we're gonna take the right point. So therefore the Raymond some is equal to the limit. As an approaches infinity, we have the limit as X approaches infinity off this sum from eyes equal to one upto end off f of X I, which is equal to one plus nine. I divided by n minus four l end of one plus nine. I divided by end as this is all multiplied by nine over and

Okay, This question wants us to find the following live it. So, as we should always do, we should try to just plug it. So our first step should be to plug in. So we get 1/0 times the integral from 0 to 0 of original expression. But the integral over with zero is also zero. So we end up getting zeroed about by zero. So we have an indeterminant form, which means we can apply low petals rule. So what we're going to do is instead take the limit as X approaches zero of the derivative of the top, which the derivative is the integral expression divided by the derivative of the bottom, which should be the derivative of execute. So we can evaluate. Each is so the top we see that won't we just use our normal derivative of an integral rule? So it's just X squared over X to the fourth plus one, and we don't have to worry about any chain growth factors. And then we divide that all by three x squared, or if we want to simplify our fraction of it, we can write this as X squared over X to the fourth plus one and then times one over three x squared. So now we can start simplifying some or and get the limit as X approaches zero of X squared over three x to six plus three x squared, and we're gonna keep getting 0/0. So let's just keep taking derivatives. So we get two x over 18 x to the fifth plus six x and again, nothing here. So we have to do another low p tall and get to over 90 x to the fourth plus six. And now we actually get somewhere because if we plug in zero, we just get to over zero plus six and to over six is just one third, and that's our answer.

Hello. So here we have the limit as X tends to infinity of one over X times the integral from zero to X. Of E to the T square Bt. Well, this is going to be just equal to the limit as X tends to infinity. Um Just by the fundamental film here, we just have, it's going to be just equal to E to the X squared um over one. And then as X goes to infinity, we just have here um E to the infinity. What's going to be equal to infinity? So therefore our limit here is equal to infinity.

Given that f of X is equal to X to the negative four thirds power. Were asked to evaluate the improper integral from one to infinity. So we can rewrite the instagram from one to infinity and substitute in our function. Now we can integrate using our general power rule for integral. So we do this your last twist X to the negative 4/3 plus three thirds, divided by negative four thirds plus three thirds, Evaluated that infinity and one. And we work this out. We are left with negative three X To the -1 3rd. And in fact I'm actually gonna rewrite this, just going to move into the bottom and next to the one third the Queudrue defects. And this can be evaluated at infinity and one. So if you plug in these numbers here We have negative three divided by the cube group of infinity minus negative three over The Cube Root of one. So negative three divided by infinity. It's just gonna go to zero in minus a negative three divided by one. The negatives are going to get the gate themselves and we're left with a positive. So the final answer is three


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