L'Hôpital's Rule

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take x plus sine x, over x, as x tends to infinity. The top and the bottom both tend to infinity. So a student uses today's rule: differentiate the top, differentiate the bottom, and gets one plus cos x, over one. But cos x keeps swinging between minus one and one. So the student writes: no limit. Is that right? Hold that thought. Today's topic is L'Hôpital's rule, a fast way to handle zero over zero and infinity over infinity limits. It comes up often in JEE, so let's begin.

0:42When can we use L'Hôpital's rule?

So, we have a fraction, and x tends to a point a, or to plus or minus infinity. That is the plus minus sign before the sideways eight, the infinity symbol, okay? Condition one: the form must be zero over zero, or infinity over infinity. Clear? Condition two: f and g, the top and the bottom, must be differentiable near a, okay? Of course: this rule is all about derivatives. Right? Condition three, the one students forget: the limit of f prime of x over g prime of x must exist. The little dash is called prime, so f prime of x is the derivative of f. And lim is short for limit. Okay? When all three hold, the limit of f of x over g of x equals the limit of f prime of x over g prime of x. Three conditions, one equation. With me?

1:34Top and bottom, separately

Now, the most important point. We have a fraction to differentiate. So, do we use the quotient rule? No, no, careful! We differentiate the top on its own, and the bottom on its own. Completely separately. Got it? Out of habit, many students use the quotient rule on f of x over g of x. That gives a wrong answer. This is the classic trap. Don't fall for it! Okay? And after each round, check the form again. Still zero over zero, or infinity over infinity? Then apply the rule again. Repeat only while the form stays. Clear?

2:12Growth at infinity

Now let x tend to plus infinity; the arrow means tends to. Log x, the natural log, written l n, tends to infinity. So does x. So in log x over x, which one wins? Let's ask the rule. Infinity over infinity, so differentiate the top and the bottom: one over x, over one. That is one over x, which tends to zero. So x wins, completely! Right? Next, x cubed over e to the power x. Also infinity over infinity. One round: three x squared over e to the power x. Still infinity over infinity. Again: six x over e to the power x. And again: six over e to the power x. Three rounds, and it tends to zero. With me? So remember this order, for any positive n. Log x is much smaller than x to the power n, which is much smaller than e to the power x; the double less than sign means much smaller. So log x over x to the power n tends to zero, and x to the power n over e to the power x tends to zero. Okay?

3:180 times infinity form: rewrite as a fraction, x ln x

Now, the rule works only on a fraction. So what about zero times infinity, or infinity minus infinity? The dot means times. Rewrite them as a quotient first, then use the rule. Got it? Course example: x log x, as x tends to zero from the right; the small plus sign after the zero means from the right. Here x tends to zero, and log x tends to minus infinity. Zero times infinity. So how can we write it as a fraction? Yes! Send x to the bottom as one over x: log x, over one over x. Now it is the infinity over infinity form, shown in brackets. With me? Differentiate separately: one over x, over minus one over x squared. That is minus x, which tends to zero. So x log x tends to zero.

4:11When the rule says nothing

Clear? Now, back to our puzzle. x plus sine x, over x, as x tends to infinity. Infinity over infinity form, yes. Right? Differentiate separately: one plus cos x, over one. It oscillates; it keeps swinging and never settles, so it has no limit. So the original limit does not exist either, correct? No, no, wait! Look at condition three: the limit of f prime over g prime must exist. Here it does not. So the rule simply gives no information. It does not say no limit; it says nothing at all. Clear? So, go directly. Divide by x: one plus sine x over x. Sine x stays between minus one and one, so sine x over x tends to zero. The limit is one! Puzzle solved: our student fell into the trap!

5:04Example 1: limit of (x - sin x)/x cubed as x tends to 0

Alright, example one, from the course. Use L'Hôpital's rule to find the limit as x tends to zero, of x minus sine x, over x cubed. Pause the video, and try it first. Done? Okay, let's check. As x tends to zero, the top and the bottom both tend to zero. Zero over zero form, so the rule applies. Differentiate separately: the top becomes one minus cos x, and the bottom becomes three x squared. With me? Now, before you substitute: what form is this new fraction? Zero over zero again, exactly! One minus cos zero is zero. But we know the standard limit: one minus cos x, over x squared, tends to one half. So we get one third times one half: one sixth. Okay? Or, if you forget that standard limit, apply the rule once more: sine x over six x. Sine x over x tends to one, so this tends to one sixth again. The answer is one sixth, both ways. That's it!

6:13Example 2: limit of ln sin x over (pi - 2x) squared at pi/2

Now a slightly harder one, JEE level. Find the limit as x tends to pi over two, of log of sine x, over pi minus two x, whole squared. Pause and try. Hint: be careful with the chain rule at the bottom. Done? Okay, let's check. First, the form. At pi over two, sine x is one, and log one is zero. At the bottom, pi minus pi is zero. Zero over zero, so the rule applies. Right? Now differentiate the bottom first. What is the derivative of pi minus two x, whole squared? Two times pi minus two x, times minus two, by the chain rule: minus four times pi minus two x. Oh no, careful! Many students forget that minus two. And the top: log of sine x gives cos x over sine x, that is cot x. At pi over two, cot x is zero and the bottom is zero. Still zero over zero. With me? So apply the rule once more. The derivative of cot x is minus cosec squared x, and the derivative of minus four times pi minus two x is plus eight. At pi over two, cosec x is one, so we get minus one over eight. The answer is minus one eighth. Got it?

7:33Quick revision: conditions, separate derivatives, growth

So, quick revision. One: use the rule only for zero over zero or infinity over infinity, with f and g differentiable, and the limit of f prime over g prime existing. Two: differentiate the top and the bottom separately, and repeat while the form stays. Three: for zero times infinity or infinity minus infinity, make a quotient first. And remember, for positive n, log x is much smaller than x to the power n, which is much smaller than e to the power x. Now go and try the practice problems on this page. All the best!