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Q1k Find the limit for each of the following functions if is exist ?1) limlx - 3| x-32) IF f(x) = {x+2 ~1<7< 0 find limf(x) X , 0 < x < 1 X-0Q22x2_9 X+3...

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Q1k Find the limit for each of the following functions if is exist ?1) limlx - 3| x-32) IF f(x) = {x+2 ~1<7< 0 find limf(x) X , 0 < x < 1 X-0Q22x2_9 X+3 -6A) f(x) X #-3 UNAVERvT X =-3 Is f cont: at x=-3 ? 51 x2_8 B) f(x) = X+3 x <6 -6 6 < x < 1 Is f cont: at X-6 ? Ianl O3zFind domain and range for the following functions? fx + 2 -1< * < 0 1) fx) = x2_1 2) f(x) = X _ 0 < x < 1

Q1k Find the limit for each of the following functions if is exist ? 1) limlx - 3| x-3 2) IF f(x) = {x+2 ~1<7< 0 find limf(x) X , 0 < x < 1 X-0 Q22 x2_9 X+3 -6 A) f(x) X #-3 UNAVERvT X =-3 Is f cont: at x=-3 ? 51 x2_8 B) f(x) = X+3 x <6 -6 6 < x < 1 Is f cont: at X-6 ? Ianl O3zFind domain and range for the following functions? fx + 2 -1< * < 0 1) fx) = x2_1 2) f(x) = X _ 0 < x < 1



Answers

Find the limit of each function (a) as $x \rightarrow \infty$ and (b) as $x \rightarrow-\infty .$ (You may wish to visualize your answer with a graphing calculator or computer.) $$f(x)=\frac{2}{x}-3$$

We want to find the limit of rational function which is F of X equals React to the 7th. Yeah plus five X squared minus one. Okay. Which is divided by six X cubed minus seven X plus three. Yeah. So we see that the limit as X approaches infinity will be infinity because this will grow infinitely. And this term here is going to grow infinitely but it's a positive term. So as X goes to infinity we're going to infinity. And then as X goes to negative infinity, this function will also go to negative infinity because now this term will be negative but this term will also be negative um and it's going to grow infinitely based on the degree. So that's our final answer.

Given the situation we express support seven or you know there's work, so we have to find the equation at Lynette's X on approach. Infinity and axeman approaches minus infinity So force. Let's just, uh, modify it. Divide the numerator and denominator by express to Part seven that be three plus five, divided by X based two Port five minus form by X rays to power seven. 200 by 600 by expiration Par four minus seven Divided by X rays are sex plus three divided by expressed for seven. Now defined the limits Me. Just replace the X with infancy here that should then become three plus five bucks five by infinity, minus one by infinity. Fired by six by infinity minus 75 infinity plus three by infinity. I'm not using the powers here because infinity and the powers they made the same strength in a race apart. Anything is infinitely itself. Um, and since all summer divide by infinity, they become or approach zero. So this would become three plus 0 to 10 The answer becomes infinity Now for minus infinity video. Same in a place all of thes minus infinity. So don't bathe three plus 5211 minus and ability minus one. Divide by minus infinity. Fun by six to have a minus infinity minus 700 by minus infinity plus 3 to 1 by minus Infinity. Now, the result would be same like, because thes are infinities, so will be super. Here we should become three plus zero those 00 which is again infinity. So the answer is infinity.

In this context Caution. Combining what parts up there and we weaken like a makeup extent to pass minus infinity X plus one Maxwell plus three. Let me guess you may not extra stop last menacing Lindy Dividing excluding the American denominator. Have one. My ex boyfriend, my ex quipped, all divided by one blessed. Now we're extends to plus or minus Infinity dictum tends to zero. This time tends to zero and they stumble. Second studio. So we're left with zero last zero divided by unless it

For the following problem, We want to consider the limit as X approaches three from the right, the left, And actually at the value knowing we have X -1 times X -2 over X -3. So we see that as we approach three from the left or from the right, this number will be positive, this number will be positive and this number will be positive. So approaching only positive numbers, meaning that this is going to be infinity again, because this value is getting extremely small, where these values are larger compared to this tiny value, then we're going to consider as we approach it from the left. So if we approach from the left, this value will be negative, this value will be positive and this value will be positive. So we end up getting a larger positive value over a very small negative value, making this negative infinity. So the limit does not exist.


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