Example: lim x→ 0 sin x x2 If we apply l’Hôpital’s rule to this problem we get: lim x→0 sin x x2 cos x 2x − sin x = lim x→ 0 2 = 0. = lim x→ 0 (l’Hop) (l’Hop) If we instead apply the linear approximation method and plug in sin x ≈ x, we get: sin x x2 ≈ ≈ x x2 1 . x We then conclude that: sin x x2 sin x lim x→0− x2 lim x→0+ = ∞ = −∞. There’s something ﬁshy going on here. What’s wrong? Student: L’Hôpital’s rule wasn’t applied correctly the second time. 1 x 0 That’s correct; limx→0 cos 2x is of the form 0 , not 0 or some other indetermi nate form. This is where you have to be careful when using l’Hôpital’s rule. You have to verify that you have an indeterminate form like 00 or ∞ ∞ before applying the rule. The moral of the story is: Look before you l’Hôp. Also, don’t use l’Hospital’s rule as a crutch. If we want to evaluate: x5 − 2x4 + 1 x→∞ x4 + 2 we can apply l’Hôpital’s rule four times, or we could divide the numerator and denominator by x5 to conclude: lim x5 − 2x4 + 1 x→∞ x4 + 2 lim 1 − 2/x + 1/x5 x→∞ 1/x + 2/x5 1 = 0 = ∞. = lim After enough practice with rates of growth, we can calculate this limit almost instantly: x5 x5 − 2x4 + 1 ∼ lim lim = ∞. x→∞ x→∞ x4 x4 + 2