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MAT237Y1 (53)
Lecture

4.4 lhopitals.pdf

4 Pages
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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Fri 10/10/08 4.4 - L’Hopital’s!!! And now probably the moment you’ve all been waiting for! Let’s do some L’Hopital’s! ...it certainly makes taking limits much much easier! To begin, let f(x) and g(x) be di▯erentiable functions where f(a) = g(a) = 0. In other words, if we try to take the limit, f(x) 0 lim = : x!a g(x) 0 So the idea is, we can’t really take the limit right now. We may be able to factor out something, but not necessarily (like, say we had sin(x)=x). Instead, we’re going to dabble a little in section 3.8 Linear Approximation. Remember that stu▯? We brie y went over it earlier this week. So, given f a di▯erentiable function, then for x values that are pretty darn close to a, f(x) ▯ f(a) + f (a)(x ▯ a) More importantly, if we consider the limit as x ! a for f(x) and g(x), then 0 x!a f(x) = lx!af(a) + f (a)(x ▯ a) and 0 x!am g(x) = x!a g(a) + g (a)(x ▯ a) So, let’s take the limit of the quotient and see what happens! 0 lim f(x) = lim f(a) + f (a)(x ▯ a) x!a g(x) x!a g(a) + g (a)(x ▯ a) And, well, f(a) = 0 and g(a) = 0 (we assumed this above), so we get 0 f(x) f (a)(x ▯ a) x!a = lx!a 0 g(x) g (a)(x ▯ a) Finally, we can just cancel out the (x ▯ a)’s, and we’re just left with f(x) f (a) lim = lim 0 x!a g(x) x!a g (a) And since f(x) and g(x) are nice and di▯erentiable (because we assumed they are), then lim f (x) = f (a), and lim g (x) = g (a). Hence, x!a x!a 0 f(x) f (x) x!a = x!a 0 g(x) g (x) Yay! This is L’Hopital’s Rule! And guess what! It doesn’t just have to be limx!at can be one-sided 0 limits, or even limits to in▯nity! The only thing that you need is the indeterminate 0 . We’ll do more cases of L’Hopital’s later, but ▯rst, i want to recap what we just did: L’Hopital’s Rule: If f and g are di▯erentiable functions and lim f(x) = 0 x!a g(x) 0 1 then f(x) f (x) lim = lim 0 x!a g(x) x!a g (x) 0 In other words, if when you try taking a limit, you get the indetermina0e, go ahead and take the derivative of the top and bottom separately, and then try the limit again! Notice this is not like the quotient rule! We’re not taking the derivative of the entire thing here; we’re treating and keeping f(x) and g(x) as two separate functions. Example: ▯nd 2 x ▯ 1 x!1m x ▯ 1 Well, what we’ve always done before is factor out the (x ▯ 1), which is what’s causing the problem in this limit. So, x ▯ 1 (x + 1)(x ▯ 1) lim = lim = lim x + 1 = 2 x!1 x ▯ 1 x!1 x ▯ 1 x!1 0 Now, doing this same limit using L’Hopital’s, ▯rst we need to check that it’s of the fo0. Remember, in order for L’Hopital’s to work, it HAS to be of the form 0!!! Plugging in x = 1 0 certainly con▯rms this! So, x ▯ 1 2x lim = lim = 2 x!1 x ▯ 1 x!1 1 Example: Here’s one we’ve used many many times in class: sin(▯) lim ▯!0 ▯ First, check is it? yup! Let’s do it! 0 sin(▯) cos(▯) lim = lim = 1 ▯!0 ▯ ▯!0 1 Example: How about this one? sin(9x) 9cos(9x) 9 x!0 sin(3x) = x!0 3cos(3x) = 3 = 3 Example: x
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