Continuity at a Point

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take f of x equals x squared minus one, over x minus one. At x equals one, it gives zero over zero: no value at all. Everywhere else, it is just the straight line x plus one, so the graph is a line with one tiny hole. Is one missing point enough to break the graph? And if I fill the hole with the value three, is it whole again? Hmm. Our eyes cannot decide this; we need a precise test. That test is continuity at a point, and it comes up often in JEE, okay? So, let's begin.

0:41Continuity at x=a: three checks

So, what does continuous at a point really mean? Imagine walking along a graph towards the point x equals a, okay? You can arrive from the left, or from the right, and then there is the value at the point itself. So tell me: which three numbers must match? Yes, exactly! The limit from the left, the limit from the right, and the value f of a, at the point itself. On screen, lim means limit, and the arrow means tends to. The small minus sign on a means from the left; the small plus sign, from the right. So, f is continuous at x equals a if the limit as x tends to a from the left, equals the limit as x tends to a from the right, equals f of a, okay? Now, three checks. Check one: f of a is defined. The function must actually have a value there. Right? Check two: the limit as x tends to a of f of x exists. There is no small sign on a, so we come from both sides. And this limit exists only when the left and right limits are both finite and agree. Clear? Check three: the limit equals f of a, the value. Value, limit, equal. If even one check fails, f is discontinuous there, okay? And never stop after check two. You will see why soon.

2:07One side at a time

Sometimes we look at only one side. f is right-continuous at a if the limit as x tends to a from the right equals f of a, with me? Left-continuous is the same idea from the left. Now think: if a function is both left-continuous and right-continuous at a, what is it? Continuous at a, correct! Both together is exactly our definition. Got it?

2:33When continuity breaks: four types

Alright, what can go wrong? The course lists four types of discontinuity. First, removable: look at the left graph. The open circle is a hole, and the curve heads straight into it, so the limit exists. But the value, the orange dot, sits somewhere else, or sometimes it is missing. So, how would you repair this graph? Simple! Redefine f of a as the limit. Move the dot into the hole, and the break is gone. That is why it is called removable. Okay? Second type: the jump, in the middle graph. LHL means left hand limit, and RHL means right hand limit. Both are finite, but unequal, so the graph jumps. Can a new value of f of a remove this one? No, no! One value cannot sit on two levels. So the jump is non-removable, a discontinuity of the first kind. Its size is the absolute value of RHL minus LHL; the vertical bars mean absolute value. Clear? Third type: infinite, the right graph. Here a one-sided limit is plus or minus infinity, like one over x at zero: up on one side, down on the other. Right? And fourth: oscillatory. Like sine of one over x at zero: it keeps swinging up and down, faster and faster, and never settles. With me?

4:03Example 1: is (x2 - 1)/(x - 1) continuous at x = 1?

Now, remember our puzzle? Here it is. f of x equals x squared minus one, over x minus one, for x not equal to one; the slashed equals sign means not equal to. Is f continuous at one if, part a, f of one is two? And part b, f of one is three? Pause the video, and run the three checks yourself. Done? Okay, let's check. x squared minus one is x minus one, times x plus one. Cancel x minus one; that is allowed, because x is not equal to one. So f of x is just x plus one. Okay? Now, as x tends to one, x plus one tends to two, from both sides. So the limit is two, in both parts. Right? Part a: f of one is two, and the limit is two. All three checks pass: continuous! Part b: f of one is three. The value is defined, and the limit exists. So, continuous? No, careful! Check three fails: three is not equal to two. The limit exists but the value is wrong, so it is a removable discontinuity. Redefine f of one as two, and it is repaired. So, our puzzle: with no value at one, check one fails, and filling the hole with three does not fix it either. Only the value two makes the graph whole. Clear?

5:33Example 2: continuity of x[x] at x = 0 and x = 1

Now a harder one. Discuss the continuity of f of x equals x times the greatest integer less than or equal to x, at x equals zero and x equals one. The square brackets stand for that greatest integer; we will just say greatest integer of x. Pause the video and try it. Hint: take each side separately. Here is a quick answer many students give. Greatest integer of x jumps at every integer, so f must break at zero and at one, right? Not so fast! Never guess; calculate. The sign with a line under it means less than or equal to, and the plain one means less than. For x from minus one up to zero, greatest integer of x is minus one; from zero up to one, it is zero; and from one up to two, it is one. Okay? Now at zero. Left hand limit: just left of zero, f of x is x times minus one; the dot means times. That tends to zero. Right hand limit: x times zero, which is zero. And f of zero is zero. With me? All three equal, so f is continuous at zero! The greatest integer jumps there, but the factor x tends to zero and flattens the jump. So the quick answer was wrong. See the trap? Now at one. Just left of one, greatest integer of x is zero, so the left hand limit is the limit of x times zero: zero. Just right of one, it is one, so the right hand limit is the limit of x times one: one. Got it? Zero and one do not agree, so the limit does not exist. f is discontinuous at one: a jump, of size one. Continuous at zero, discontinuous at one. Always check each side, and never trust the quick answer!

7:33Quick revision: three checks, removable and jump

So, quick revision. Continuous at a: three checks, left hand limit equals right hand limit equals f of a, all three equal. Removable: the limit exists, so redefine f of a as the limit. Jump: the two sides differ, and the size is the absolute value of their difference. Now go and try the practice problems on this page. You can do them. All the best!