Question
Determine the limit by sketching an appropriate graph:1-X0sX < 1lim f(x) , where f(x) =1sx < 4 X=4Da The Iimit does not exist
Determine the limit by sketching an appropriate graph: 1-X 0sX < 1 lim f(x) , where f(x) = 1sx < 4 X=4 Da The Iimit does not exist
Answers
Sketch the graph of the function to find the given limit, or state that it does not exist. $\lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{2}-1}$
Okay, so we want to find are following, um, limit belly. So looks notice. Stop you in a numerator. We can actually fact Curtis into an expose four. And then what we have remaining. That's going to be a two x minus two. So here we have two x squared, minus a four on. Then let's see two x squared, minus two x and then plus each eggs. Actually, since there's a four here should be a one. Okay, so that's good. I know. Let's cancel out our like terms and how we can use drugs up. So that's two times negative for minus one. So that's equal to negative AIDS minus one. So we hit our limit. Value is equal to negative night.
Okay, so we want to simplify or find a following limit. So we want to start by factory each them. So this year we can rewrite this as a difference of two squares. We get X squared minus one times x squared plus one. And here we can factor this into a X squared. Um, that's going to be minus for and in an X squared plus one. Okay, so I would cancel out our like terms. And now let's used Excel. So that's going to be one squared minus one over once. Word what it's for. So that's going to give us a zero over negative three. So or limit valley here is equal to dough.
To sketch the graph to determine if this limit exists or not before we do. Because when you sketch this graph here, going to find out that it's going to be straight line, some looking something like this, except there's a whole right here. So that specific coordinate is actually the limit that we're looking for. And so the reason why we know this is because if we factor out X squared minus one, that's X minus one times X plus one, any time we have a rational function and we can eliminate a factor top and bottom. And the resulting expression you can take and apply with this, no matter basically have a whole. And that means that the limit does exist there because we can approach it from both sides. That limit as X approaches one is one plus one she goes to. So therefore the limit here is to that's one that's two here. So the limit exists.
Really given problem, we want to sketch the graph of the function to find the given limit Or state that it doesn't exist. So here, we're gonna be looking at the limit as X approaches zero of one plus one over X. Mhm. So this is one of those cases where if we were to just plug in zero immediately, it wouldn't work because we'd get undefined value. So in this case the graph is actually a very helpful way for us to determined The actual value of the function or the limit. So we see that as X goes to zero from the right, we end up getting infinity and as X goes to zero from the left, we end up telling negative infinity. So in this case we see that as a result this limit is going to be undefined.