148 CHAPTER 5. THE DERIVATIVE

Warning 5.17.5 Always check assumptions of L’Hôpital’s rule before usingit.

Example 5.17.6 Find limx→0+cos2x

x .

The numerator becomes close to 1 and the denominator gets close to 0. Therefore, theassumptions of L’Hôpital’s rule do not hold and so it does not apply. In fact there is no limitunless you define the limit to equal +∞. Now lets try to use the conclusion of L’Hôpital’srule even though the conditions for using this rule are not verified. Take the derivativeof the numerator and the denominator which yields −2sin2x

1 , an expression whose limit asx → 0+ equals 0. This is a good illustration of the above warning.

Some people get the unfortunate idea that one can find limits by doing experiments witha calculator. If the limit is taken as x gets close to 0, these people think one can find the limitby evaluating the function at values of x which are closer and closer to 0. Theoretically,this should work although you have no way of knowing how small you need to take x toget a good estimate of the limit. In practice, the procedure may fail miserably.

Example 5.17.7 Find limx→0ln|1+x10|

x10 .

This limit equals limy→0ln|1+y|

y = limy→0

(1

1+y

)1 = 1 where L’Hôpital’s rule has been

used. This is an amusing example. You should plug .001 in to the functionln|1+x10|

x10 and seewhat your calculator or computer gives you. If it is like mine, it will give 0 and will keep onreturning the answer of 0 for smaller numbers than .001. This illustrates the folly of tryingto compute limits through calculator or computer experiments. Indeed, you could say thata calculator is as useful for understanding limits as a bicycle is for swimming. Those whosay otherwise are either guilty of ignorance or dishonesty.

There is another form of L’Hôpital’s rule in which

limx→b−

f (x) =±∞ and limx→b−

g(x) =±∞.

Theorem 5.17.8 Let [a,b]⊆ [−∞,∞] and suppose f ,g are functions which satisfy,

limx→b−

f (x) =±∞ and limx→b−

g(x) =±∞, (5.14)

and f ′ and g′ exist on (a,b) with g′ (x) ̸= 0 on (a,b). Suppose also

limx→b−

f ′ (x)g′ (x)

= L. (5.15)

Then

limx→b−

f (x)g(x)

= L. (5.16)

Proof: By the definition of limit and 5.15 there exists c < b such that if t > c, then∣∣∣∣ f ′ (t)g′ (t)

−L∣∣∣∣< ε

2.