Sketch a graph of f and use it to make a conjecture about the ...

Question

Sketch a graph of f and use it to make a conjecture about the values of f(a), lim x→a^−f(x),lim x→a^+f(x), and lim x→a f(x) or state that they do not exist.

f(x) = x^2−25 / x−5; a=5

Accepted Answer

Welcome back, everyone. In this problem, we're analyzing the function HX given by H X equals X2 minus 16 divided by X minus 4, where A equals 4. Sketograph of H and determine the values of H of A. The limit as x approaches A from the left of H of X. The limit as X approaches a from the right of H of X, and the limit as X approaches A of H of X or indicate if they do not exist. A shows the graph and says that H of A does not exist. The limits from the left and the right of H of X are 8, and the limit of H X as X approaches A equals 8. B says they are 888 and the and that the limit of F of X sorry H X as X approaches A does not exist. See again shows the graph and says that they are all 8. And D OK. Says that they do not exist, that sorry, H of A does not exist and all of the limits specified equals it. And here it shows our graph. Now, for this problem, let's first see if we can get an idea of what the graph is going to look like, and then try to figure out what the values of the function and the limits will be, OK? No. We know that our function it affects. Is equal to x2 minus 16 divided by X minus 4. Now we could rewrite our numerator as x minus 4 multiplied by x + 4, the difference of two squares, and we are dividing that by X minus 4. And now we notice that X minus 4 cancels X minus 4, which leaves us with X + 4. In other words, the numerator and denominator have the common factor X minus 4. Now this canceled value creates a hole in the graph at X equals 4, OK? And now at X equals 4. OK. Then our function, our value H of 4 would be equal to 4 + 4, which equals 8, OK? But again, remember that. The functions were canceled at this value. So H of 4 does not exist. In other words, OK. Then We are really saying that the canceled value. Creates a hole, OK. In the graph At 48, OK. So now, now that we have our graph and we know that eventually it could simplify the X + 4, we can draw a line at Y equals X + 4 and then draw a hole at 4A. OK? So let's go ahead and do that. Let's just do a sketch of that, OK? I think I could do it over here. Mhm. So now we have our X and Y axis. Uh for X + 4, that means the Y intercept is 4, OK. And now, for this graph, like we said, we know that when X equals 4. OK, there's a hole in our graph. When Y equals 8, so in other words, it's going to look something like this. Now that we have the sketch, we can start to think about what the limits would be, OK? So, so far, we know, OK, that A12 A where A E equals 4 does not exist, OK? And that is, is undefined. OK. But now, when it comes to the limit, OK. The limit As X approaches A from the left of H of X, if we look at our graph, we know that from the left, the limit or the function is approaching it. Therefore, that limit is going to be equal to it. Likewise, the limit as X approaches A from the right of H X, OK, is also going to be equal to A because it is a straight line that is discontinued at the point for it or discontinuous at the point for it. Now, what about the limit, OK, as X approaches A of H of X? Well, this limit exists if the limit from the left hand side is equal to the limit from the right hand side for that value, and the answer is going to be what those limits would be equal to. Now, in this case, notice the left hand and right hand limits are equal. So this also means that the limit of the function when X approaches 4 exists and would be equal to 8. Thus, H of A does not exist and or limits that is as X approaches A from the left, the right, and as X approaches A are all equal to it. Therefore, if we look back at our answer choices, if we look back at our graphs, OK, that will tell us then. That answer choice A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Sketch a graph of f and use it to make a conjecture about the values of f(a), lim x→a^−f(x),lim x→a^+f(x), and lim x→a f(x) or state that they do not exist.
f(x) = x^2−25 / x−5; a=5

Key Concepts

Limits

Continuity

Graphing Rational Functions

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