𝑓(𝑥) 𝑔(𝑥) = 𝑒 2𝑥 −1 𝑥 ≈ 𝑓 0 + 𝑓 ′ 0 (𝑥−0) 𝑥 Lesson: _____ Section 4.7 L’Hôpital’s Rule, Growth, & Dominance Limits of quotients can often be tricky to evaluate. For example: Direct substitution yields this “indeterminate form” which basically means we can’t determine anything about the limit from this info. lim 𝑥→0 𝑒 2𝑥 −1 𝑥 = 0 0 We’ve had a few tricks up our sleeve in the past, such as canceling common factors, rationalizing, and breaking into cases, but none of those work here. 𝑓(𝑥) 𝑔(𝑥) If we thing of this expression as we can use local linearity to explore the limit near zero. 𝒚= 𝒚 𝟏 +𝒎(𝒙− 𝒙 𝟏 ) ≈ 0+2 𝑒 2 0 (𝑥) 𝑥 ≈ 2(𝑥) 𝑥 ≈ 2 1 𝑓(𝑥) 𝑔(𝑥) = 𝑒 2𝑥 −1 𝑥 ≈ 𝑓 0 + 𝑓 ′ 0 (𝑥−0) 𝑥 lim 𝑥→0 𝑒 2𝑥 −1 𝑥 = 2 As x  0, this approximation gets better and better, so

If direct substitution yields 0 0 Think about it: In general, if f(a) = 0, and g(a) = 0, we could use local linearization to rewrite the limit as follows: lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) ≈𝒇(𝒂)+𝒇′(𝒂)(𝒙−𝒂) ≈𝒈(𝒂)+𝒈′(𝒂)(𝒙−𝒂) ≈𝟎+𝒇′(𝒂)(𝒙−𝒂) ≈𝟎+𝒈′(𝒂)(𝒙−𝒂) 𝒇′(𝒂)(𝒙−𝒂) 𝒈′(𝒂)(𝒙−𝒂) ≈ = 𝑓′(𝑎) 𝑔′(𝑎) If direct substitution yields 0 0 we can still find the value of the limit by plugging into the derivatives instead! What is this telling us ????

L’Hôpital’s Rule lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = If f and g are differentiable, and 𝑓 𝑎 =𝑔 𝑎 =0, and g’(a) ≠ 0, then 𝑓′(𝑎) 𝑔′(𝑎) lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = Or, more generally, lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = lim 𝑥→𝑎 𝑓′(𝑥) 𝑔′(𝑥) Dude, that’s my rule. Get your own! See p.215 ex. 2

L’Hôpital’s Rule also applies to limits involving infinity! Provided that f and g are differentiable: If lim 𝑥→𝑎 𝑓 𝑥 = ±∞ and lim 𝑥→𝑎 𝑔 𝑥 = ±∞ or if lim 𝑥→∞ 𝑓 𝑥 = lim 𝑥→∞ 𝑔 𝑥 = 0 or lim 𝑥→∞ 𝑓 𝑥 =±∞ 𝑎𝑛𝑑 lim 𝑥→∞ 𝑔 𝑥 =±∞ We can use L’Hopital’s Rule! Then it is can be shown that lim 𝑥→𝑎 𝑓(𝑥) 𝑔(𝑥) = lim 𝑥→𝑎 𝑓 ′ (𝑥) 𝑔 ′ (𝑥) (provided that the limit on the right-hand side exists) See p.216 ex. 3,4,6 Stop it. I’m blushing.