Greatest Integer, Fractional Part and the Sandwich Theorem

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle before we start. You know that sine x over x tends to one, as x tends to zero. And tan x over x also tends to one. Now put each of them inside the greatest integer function. Same limit inside, so the same answer outside, right? Hmm. One of them gives zero, and the other gives one! How? Today we learn the tools that settle such questions: the greatest integer and fractional part inequalities, and the sandwich theorem. Questions like this come up often in JEE, so let's begin.

0:41Two tools: greatest integer [x] and fractional part {x}

First, a quick recap from Unit one. The square brackets around x mean the greatest integer less than or equal to x; from now on we just say greatest integer of x. The curly brackets mean the fractional part of x. And every real x splits as greatest integer of x, plus fractional part of x. Okay? Now, greatest integer of x is never more than x. But how far below x can it be? Less than one, exactly! So x minus one is less than greatest integer of x, which is less than or equal to x. The strict sign on the left, and the sign with a line under it on the right. This one line turns a jumpy function into something we can squeeze. Clear? And the fractional part is at least zero, and always less than one. Zero is allowed, one never. Got it?

1:38The sandwich theorem

Now our second tool, the sandwich theorem. Suppose f of x sits between g of x and h of x, for all x near a. Like the filling between two slices of bread. With me? And suppose both slices tend to the same number, capital L, as x tends to a. Lim is short for limit, and the arrow means tends to. Right? Then, where can the filling, f of x, go? Nowhere else, correct! The limit of f of x is also capital L, the same number. But check both conditions: the inequality must hold for x near a, and the two outer limits must be equal. Okay?

2:19Bounded times something tending to zero

Now a quick rule that comes straight from the sandwich. Bounded, times something that tends to zero, tends to zero; the cross means times. Bounded means it stays between two fixed numbers. Got it? Course example: x times sine of one over x, as x tends to zero. Sine of one over x keeps swinging between minus one and one. So what does the product do? It tends to zero, yes! The vertical bars mean modulus, the size. Sine of one over x is at most one in size, so the size of the product is at most modulus of x, and that tends to zero. Squeezed to zero! Clear? Now try another one from the course: cos n plus sine n, over n, as n tends to infinity. What does it tend to? Zero, perfect! The top always lies between minus two and two. Careful: these square brackets mean a closed interval, not the greatest integer. Bounded top, and n tends to infinity, so the fraction tends to zero. With me?

3:28Greatest integer [u] as u tends to n: above vs below

Now, back to our puzzle. Let u be an expression that tends to an integer n. If u comes from above, always slightly more than n, then greatest integer of u is n, so it tends to n. Right? But what if u comes from below, always slightly less than n? What is greatest integer of u then? It is n minus one, yes! So the limit is n minus one. Same target n, but the direction changes the answer. This is the trap. Okay?

3:59Puzzle solved: sin x/x versus tan x/x

Remember this picture from trigonometric limits? On the circle of radius one with centre O, triangle OAP sits inside sector OAP, inside triangle OAT; the little triangle symbol just means triangle. Their areas give half sine x, less than half x, less than half tan x. So sine x is less than x, less than tan x, for x between zero and pi over two. Clear? So, for small x, is sine x over x just above one, or just below? Just below, exactly! Divide sine x less than x by x, which is positive here. And for negative x, the ratio has the same value as at minus x, which is positive. So near zero, sine x over x lies between zero and one, its greatest integer is zero, and the limit is zero, not one! Got it? And tan x over x is just above one, on both sides of zero, so its greatest integer is one, and the limit is one. Both tend to one inside, but the answers are zero and one. Puzzle solved! Right?

5:03Example 1: limit of [x]/x as x tends to infinity

Now, Example one, an easy one from the course. Find the limit as x tends to infinity, of greatest integer of x, over x; the sideways eight is the infinity sign. Pause the video, and try it first. Done? Okay, let's check. Greatest integer of x is jumpy, so we do not handle it directly. We start with our window: x minus one is less than greatest integer of x, which is less than or equal to x. Okay? Now we want to divide everything by x. But can we divide this inequality by x, and keep the signs as they are? Yes, because x tends to plus infinity, so x is positive. Dividing by a positive number keeps the signs. So x minus one over x, is less than greatest integer of x over x, which is less than or equal to one. With me? The left side is one minus one over x, which tends to one. And the right side is just one. Right? Both slices tend to one, so by the sandwich theorem, the limit is one. That's it!

6:12Example 2: limit of x[1/x] as x tends to 0 from the right

Now a harder one from the course. Find the limit as x tends to zero from the right, of x times greatest integer of one over x. The small plus sign up top means from the right. Pause the video and try. Done? Okay, let's check. Many students say: x tends to zero, so x times anything tends to zero, right? Oh no, careful! Our rule needs the other factor to be bounded. But greatest integer of one over x is not bounded; it keeps growing as x tends to zero from the right. So use the window with one over x in place of x: one over x minus one, is less than greatest integer of one over x, less than or equal to one over x. Clear? Now multiply by x. Here x tends to zero from the right, so x is positive, and the signs stay. One minus x, is less than x times greatest integer of one over x, which is less than or equal to one. Okay? As x tends to zero from the right, one minus x tends to one, and the right side is one. Sandwiched! So the limit is one, not zero. Very good!

7:27Quick revision: [x] bounds, direction trap, bounded times 0

So, quick revision. One: x minus one is less than greatest integer of x, less than or equal to x, and the fractional part lies from zero up to, but not including, one; use these to build a sandwich. Two: when u tends to an integer n, greatest integer of u tends to n from above, and to n minus one from below. Three: bounded, times something that tends to zero, tends to zero. Now go and try the practice problems on this page. You can do them. All the best!