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Use the graph to find the limit.(a) $lim _{x ightarrow-2} h(x)$(b) $lim _{x ightarrow 0} h(x)$...

Question

Use the graph to find the limit.(a) $lim _{x ightarrow-2} h(x)$(b) $lim _{x ightarrow 0} h(x)$

Use the graph to find the limit. (a) $lim _{x ightarrow-2} h(x)$ (b) $lim _{x ightarrow 0} h(x)$



Answers

Evaluate the limit, if it exists.
$ \displaystyle \lim_{h \to 0}\frac{\frac{1}{(x + h)^2} - \frac{1}{x^2}}{h} $

Limit as H goes to zero of X plus h squared minus X squared all over H. We can't do our direct substitution yet, so, uh, we're gonna have to do some sort of algebra to simplify this. Thea only step that we could make here is to expand the expos H all squared. So x squared plus two x h plus h squared minus X squared all over h and ah, the ex squares on the negative x squared that cancels out. We're left with limit as each goes to zero on the top of gonna take out a common factor of H. Well, uh, we're left with two x plus h all over h again. Another cancellation occurs here. That's, um now limit as H goes to zero of two X less each. And we can do our direct substitution now. So when we put zero in for H, we have two x

Ankles are square minus X squared, divided by arch When much goods your zero there noticed that and we trying to good as they go into this archer and his archery, which again is only Buddy Bizot, which is under fight from. Therefore we try job expand this grand describing here and wish you no limit on the much faster zero for the square, which Linda X squared plus two x plus the square when it's x squared, divided by arch and the same goes on the X Going with the X Square and then we have left with the two x plus the square leaving a but munch on now on the tongue was still confronted a hunch our side and we should get now and then to explicit art in money. But aren't we? And we can cancel that much about the art. And then we have left with the limit on this year zero to express crunch wear able to pluck in desert this arch near. Then we have in court, you know, two x doesn't immediate answer

Okay, so we want to find are falling limits as X approaches. Negative, too. So let's use decks up. So that's native to to depart to minus three times negative, too overnight of two. So that's equal to four plus 6 11 against to which is you. Put the 10 over negative, too, so that's equal to negative five. And it might be we can look at her fallen graph looking at a graph as X approaches zero from the right and the left. You see that this approach e negative frame and report see, we can use to excel again. And so that's three squared minus three times three over there in shock people to nine minus 9/3. So that gives us that we have a zero.

Okay, This question asks us about this piece, wise function and certain limits. So the first part wants us to find the limit as X approaches zero from the positive direction of H of X. And if we're approaching zero from the positive direction, it means were just slightly above zero. So it behaves like that second piece because we're greater than zero, but less than two obvious. So it's just the limit. As X approaches zero of X squared, which is zero in Part, B wants us to determine if the overall limit at zero exists. So first, let's check to make sure the negative from limit lines up. And if it does, our limits also gonna be zero. So over approaching zero from the negative direction, that means we're gonna be dealing with that ex piece because we're in the negatives. And if we do that, that's just equal to zero. And since we're both the right and the left hand limits or zero, the overall limit is equal to zero. And if that zero for our second limit was anything else, we would not have an existing limits. Then Part C. It wants the limit as X approaches one of h of X, and we don't have any issues to deal with here because we're not changing pieces or anything, and we have continuous functions on each piece. So it's just a matter of plugging in one to the peace were at, which is X squared. In that case, then Part D asks us about that second piece and if the limit exists at that switch. So we're looking for the limit as X approaches to from the minus direction first and over, approaching two from a value slightly less than it, we're still gonna be operating in the X squared piece, so that would be four then for the limit as Ex purchased two from positive direction. That's just the limit sex purchase to of our positive piece eight minus X, and that's equal to six. And now you can see that six is not equal to four. So that means our overall limit does not exist, and it does not exist because the left hand limit does not equal the right, and that's all you have to say.


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