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DM
Numerade Educator

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Problem 6 Medium Difficulty

For the function $ h $ whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.

(a) $ \displaystyle \lim_{x\to -3^-}h(x) $
(b) $ \displaystyle \lim_{x\to -3^+}h(x) $
(c) $ \displaystyle \lim_{x\to -3}h(x) $
(d) $ h(-3) $
(e) $ \displaystyle \lim_{x\to 0^-}h(x) $
(f) $ \displaystyle \lim_{x\to 0^+}h(x) $
(g) $ \displaystyle \lim_{x\to 0}h(x) $
(h) $ h(0) $
(i) $ \displaystyle \lim_{x\to 2}h(x) $
(j) $ h(2) $
(k) $ \displaystyle \lim_{x\to 5^+}h(x) $
(l) $ \displaystyle \lim_{x\to 5^-}h(x) $

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Video Transcript

mm hmm. So in this in this problem were given this graph for H of X which has these discontinuities in it and the section that's like A sign of one over X. Um it's just all so it's back and forth between four and two as it gets closer to five there and were asked a series of limits here. So let's look at the first one limit of H of X as X approaches minus three minus that modest means from the left. So that's this. So that goes higher and higher and higher up to the value of four, doesn't it? So this is for the next one is the limit as of H of X as X approaches minus three from the right. Okay, so that's this this direction now Which goes higher and hired up to that four right there. Okay, then we're asked to limit of H of X as X approaches minus three. Well, Since from the left it was four and from the right it was four Then that means that the limit approaching -3 from either direction is four. Okay, while h at -3, evaluated at -3 does not exist, does it? Because it is that open circle right there? Where there is not a value and there's not a point shown anywhere on the graph for it either. Okay, Next one is the limit of this function as x approaches zero from the left, so we're coming down this curve right here as we go to zero. So that is one. Okay, Next we have the limit as HH of X approaches zero from the right, the plus means from the right, that's down this curve and that's going to minus one. So then the limit as H of X approaches at zero is X approaches zero of H of X. Well, the problem here is from the left, it's it's positive one. And from the right, it's negative one. And so this does not exist because you have to have the same limit coming from both sides in order to have that value. And H at zero. Well, On our graph, it's a solid.up here at positive one. So that's the value of the function at zero. Okay, next we have I. Which is the limit of this function as X approaches the approaches to Well, as X approaches to from the left, We get the value of two and from the right We get the value of two, don't we? Coming in there? So that makes this too An H at two again is an open circle and there's no other dot for it anywhere else presented. And so this does not exist. Okay, make sure that looks good there. There we go. All right now, Question K. The limit of our function as x approaches five from the positive side. So that's from the right, Where's 5? At five is right here. So that's us coming down on this curve, which Is approaching three, isn't it? On the curve there. So this is three. And then the limit as x approaches five from the left. Well it's hard to tell on our curve, but it actually as we get closer and closer to five what's happening from the left over, we just keep on playing faster and faster and faster between two and four, so we never settle on a single number. Therefore that one does not exist for us. That limit does not exist and there is all of the limits that were asked for on this graph.

DM
Oklahoma State University
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