Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take sine of two x, over e to the power three x, minus one. Put x equals zero. On top, sine zero is zero. Below, e to the power zero is one, and one minus one is zero. Zero over zero. So is the answer zero? Or one? Or nothing at all? Now try a tiny x, say one ten thousandth, on a calculator. You get about zero point six six six six. Where does that number come from? By the end of this class, you will get it exactly, in a few lines. This is Part A of our limits, continuity and differentiability problem bank, at JEE Main level, and limits like this come up often. So, let's begin.

0:51Part A: a 30-minute test

So, here is what Part A holds. Eight solved problems, all at JEE Main level. Short, quick questions, okay? And they are mixed: limits, continuity, derivatives, tangents, maxima and increasing functions, all together. Nobody tells you which chapter a question comes from, just like the real exam. Right? The plan: finish Chapters one to eight, then attempt Part A as one thirty minute test. Today, two tools for its limit problems, then two of those problems, one easy and one harder. With me?

1:28Tool 1: standard limits as x tends to 0

Tool number one: the standard limits. The limit as x tends to zero, of sine x over x, equals one. The word lim with the little arrow under it is read as the limit as x tends to zero. And x must be in radians, okay? Two more. The limit as x tends to zero, of e to the power x, minus one, over x, is also one. And for a base a greater than zero, the limit of a to the power x, minus one, over x, is log a. The l n you see on screen is the natural log, log to the base e. Clear? Now the real power. In each of these, x can be replaced by any expression u that tends to zero. Sine u over u tends to one, e to the power u, minus one, over u tends to one, and a to the power u, minus one, over u tends to log a. Okay? But careful with that u. Take sine two x over x. It looks just like sine x over x. So it tends to one, right? No, no, careful! Under sine we have two x, so below we need the same two x. Sine two x over x equals two, times sine two x over two x; the small dot means times. Sine two x over two x tends to one, so the whole thing tends to two, not one. Same u on top and below, always. Got it?

2:57Tool 2: surds, and splitting into pieces

Tool number two, for surds. Root of one plus x, minus one. The tick-shaped sign is the square root. At x equals zero this is zero too, so in a denominator it causes trouble. So we multiply by its conjugate, root of one plus x, plus one: the same two terms, with the opposite sign in the middle. Okay? Now, what do the two brackets give when we multiply them? Yes, exactly! It is a difference of squares. The roots disappear: one plus x, minus one, which is just x. The surd is gone. With me? And one rule that holds it all together. If the limit of f is capital L and the limit of g is capital M, both finite, then the limit of f g, that is f times g, is L times M, okay? So we split a hard expression into pieces, each with a finite standard limit, and multiply. Clear?

3:55Example 1: limit of sin 2x over (e^3x - 1) as x tends to 0

Now, remember our puzzle from the start? Let's finish it. Find the limit as x tends to zero, of sine two x, over e to the power three x, minus one. Pause the video, and try it first. Hint: use tool one. Done? Okay, let's check. First, always check the form. Put x equals zero: sine zero, over e to the power zero, minus one. Zero over zero. So we must rewrite it before any value comes out. Right? Now think. Sine two x wants something below it. What? And e to the power three x, minus one, wants something below it too. What? Two x under sine two x, and three x under e to the power three x, minus one. Yes! So we write sine two x over two x, times three x over e to the power three x, minus one. We brought in a two x and a three x that were not there, so we fix it: times two x over three x. Okay? Now the limits. Sine two x over two x tends to one. The middle piece is the standard one turned upside down, so it tends to one over one, which is one. And two x over three x is just two thirds. One times one times two thirds: two thirds! And two thirds is zero point six six six, where the three dots mean the sixes go on forever. That is the number our calculator was closing in on. Mystery solved!

5:32Example 2: (2^x - 1) over (root(1+x) - 1) as x tends to 0

Now a slightly harder one. Find the limit as x tends to zero, of two to the power x, minus one, over root of one plus x, minus one. Pause the video, and try it. Hint: you need both tools. Done? Okay, let's check. The form first. At x equals zero, two to the power zero is one, and one minus one is zero. Below, root one minus one is also zero. Zero over zero again. Got it? So, the top is like a to the power x, minus one. What does it need below it? An x, yes, exactly! So we divide by x and multiply by x. Two to the power x, minus one, over x, times x over root of one plus x, minus one. The equals sign with a slash means not equal to: for x not equal to zero, x over x is one, so nothing changes. Okay? Now, the first piece. Two to the power x, minus one, over x. It looks just like e to the power x, minus one, over x. So it tends to one, right? Oh no, careful! That is the trap. The answer one is only for base e. Here the base is two, so the limit is log two. Log two, not one. Clear? Now the second piece, the surd. Use tool two: multiply top and bottom by the conjugate, root of one plus x, plus one. Below, the brackets give x, which cancels the x on top, since x is not zero. So the piece is just root of one plus x, plus one, and as x tends to zero, that tends to one plus one: two. With me? Both limits are finite, so we multiply. Log two times two: two log two. Split it, use the standard limit, rationalise the surd. Three moves, and the problem is done. See how fast that was?

7:36Quick revision: check the form, same u, conjugate

So, quick revision. One: check the form first; zero over zero needs rewriting. Two: the same u on top and below. Sine u over u, and e to the power u, minus one, over u, tend to one, but a to the power u, minus one, over u tends to log a. Three: for surds, multiply by the conjugate. Now go and attempt all eight problems on this page as one thirty minute test, then try the practice problems. You can do this. All the best!