Trigonometric Limits

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Put x equals zero in sine x over x, and you get the zero over zero form, which tells us nothing. Now put x equal to zero point zero one in your calculator. You get zero point nine nine nine nine eight, and more digits: almost exactly one. But switch to degree mode, and the same keys give zero point zero one seven four five. So which one is right? Let us find out. Trigonometric limits come up often in JEE, so let's begin.

0:37Why sin x/x tends to 1: compare three areas

So, look at this figure: a circle of radius one, centre O, with A at one comma zero. P is on the circle at angle x, in radians, and the line O P meets the tangent at A in the point T, okay? Now three shapes: the small triangle OAP, the sector OAP, and the big triangle OAT, see? The little triangle symbol on screen just means triangle. Which area is the smallest, and which is the biggest? Yes, exactly! Small triangle, inside the sector, inside the big triangle. Their areas are one half sine x, one half x, and one half tan x. Cancel the halves: sine x is less than x, less than tan x, for x between zero and pi over two. Clear? Now divide everything by sine x, which is positive here: one is less than x over sine x, less than one over cos x. Now let x tend to zero from the right. The arrow means tends to, and the small plus sign means from the right. What does one over cos x tend to? To one, yes! Because cos x tends to one. So x over sine x is squeezed between one, and something that tends to one. So sine x over x tends to one, from the right. With me? And from the left? Sine of minus x, over minus x, is the same as sine x over x, because the top and the bottom both change sign. So the left limit is also one. And a limit exists only when the left and right limits agree. Right? They agree, so the limit as x tends to zero, of sine x over x, equals one. The word lim, with x tends to zero written under it, is short for limit. And note: x in radians. Got it?

2:30The standard limits (x in radians)

Now, tan x is sine x over cos x. So what does tan x over x tend to, as x tends to zero? One, correct! Tan x over x is sine x over x, times one over cos x; the dot means times. One times one: one. Okay? Next, one minus cos x, over x squared, tends to one half. Look carefully: x squared at the bottom, not x. Write this one down. Clear? And the inverses. The small minus one up top means inverse, not one over. Sine inverse x over x, and tan inverse x over x, both tend to one. Right? Now, our puzzle. All these results need x in radians. The small circle means degrees, and x degrees is pi x over one hundred and eighty radians. So sine of x degrees, over x, tends to pi over one hundred and eighty; the wavy sign means about zero point zero one seven four five. Our degree mode number! Mystery solved. Got it?

3:43Match the angle with the denominator

Now, the real power of this result. The angle can be anything; call it u. As long as u tends to zero, sine u over u tends to one. The angle and the denominator must match. Okay? Course example: sine five x, over sine three x, as x tends to zero. What is the answer? Five over three, very good! Match each angle: sine five x over five x, times three x over sine three x, times five over three. The first two parts tend to one, so the limit is five over three. Clear? In general, sine a x over sine b x tends to a over b, for b not equal to zero; that is the slashed equals sign. With me? And tan four x over x, from the same example? Four times tan four x over four x, which tends to four. That's it!

4:42When x tends to a, not 0: put x = a + h, h tends to 0

Now, what if x tends to a number a, not zero? Put x equals a plus h; as x tends to a, h tends to zero. Then use identities to bring in sine h, tan h, or one minus cos h. Okay? Course example: the limit as x tends to pi, of sine x over pi minus x. Here, put x equals pi minus h, so h tends to zero. Then what does sine x become? Sine of pi minus h is sine h, exactly! And the bottom, pi minus x, is just h. So we get sine h over h, which tends to one. Simple, right?

5:27Products of cosines: a shortcut

One more shortcut from the course. One minus cos A cos B cos C splits into three pieces: one minus cos A, plus cos A times one minus cos B, plus cos A cos B times one minus cos C, okay? Multiply it out to check. Right? Now take the angles as a x, b x and c x, and divide by x squared. Every cos in front tends to one, and the limit is a squared plus b squared plus c squared, over two. Why squared? Example one will show you. Clear? Course example: one minus cos x cos two x, over x squared. Only two cosines here, with a equal to one and b equal to two, so there is no third term. So one squared plus two squared, over two: five over two. That's it!

6:18Example 1: limit of (1 - cos 4x)/x squared as x tends to 0

Now, Example one, an easy one from the course. Find the limit, as x tends to zero, of one minus cos four x, over x squared. Pause the video, and try it first. Done? Okay, let's check. The angle is four x, so we multiply by four, and get four times one half: two, right? Oh no, careful! The bottom must be the angle squared: four x, whole squared, which is sixteen x squared. So we multiply and divide by sixteen, not four. Clear? Now one minus cos four x, over four x whole squared, tends to one half. Sixteen times one half: eight. That is four squared over two, just like the cosine shortcut. Very good!

7:09Example 2: limit of (tan x - sin x)/x cubed as x tends to 0

Now a harder one, Example two from the course. Find the limit, as x tends to zero, of tan x minus sine x, over x cubed. Pause the video and try. Hint: take tan x common. Done? Okay. Near zero, tan x and sine x are both close to x. So the top is x minus x, zero, and the answer is zero, right? No, no, wait! The top is not exactly zero, just tiny, and the bottom, x cubed, is tiny too. So we cannot drop it. Take tan x common: tan x minus sine x equals tan x, times one minus cos x, because tan x times cos x is sine x. Okay? Now split x cubed as x times x squared. We get tan x over x, times one minus cos x over x squared. Both are standard limits! What does each part tend to? One, and one half, yes! One times one half: one half. Not zero! So when sine x and tan x are subtracted, do not replace them by x; factor first. Remember that!

8:25Quick revision: sin u/u, (1 - cos u)/u squared, x = a + h

So, quick revision. One: sine u over u, and tan u over u, tend to one as u tends to zero, u in radians. Two: one minus cos u, over u squared, tends to one half; match the angle, and square it below. Three: when x tends to a, not zero, put x equals a plus h, then use identities. Now go and try the practice problems on this page. All the best!