Lecture 17: Indeterminate Forms and l’Hˆopital’s Rule

Today: Indeterminate Forms, l’Hˆopital’s Rule

e3x − 1 Example 81 (Warm-Up). Find all values of x in [−1, 4] at which the graph of f(x)= sin(x) has a vertical asymptote. We check for VA at points where the function is undefined. For this f(x), this is exactly when 0 = sin(x), so when x =0andx = π. At x = π, we test from the right first, using the informal language that we used to describe the behavior of the function as x gets close to π but x>π:

e3x − 1 e3π − 1 lim = = −∞. x→π+ sin(x) Small −

Because one of the one-sided limits is infinite, we know that the graph of f(x) has a vertical asymptote at x = π. At x = 0, when we try to test from one side, we obtain 0 lim f(x)“=” , x→0− 0 indeterminate form of type 0 a meaningless expression. This is an example of an 0 .Ifyou come across this, you have no information about the (one-sided) limit: it could either exist, be infinite, or not exist. We’ve seen this type of thing before, though, and we used algebra to get around it. Indeterminate Forms and l’Hˆopital’s Rule In the previous example we ran into an indeterminate form when computing a limit, which is a meaningless expression that indicates we need to do additional work. There are several types of indeterminate forms, a few of which are 0 ∞ , , ∞−∞, 0 ·∞, 00, ∞0, and 1∞. 0 ∞ Luckily we have a tool for such situations. f(x) Theorem 8 (l’Hˆopital’s Rule). If f(x)andg(x) are differentiable, and if lim is an x→a g(x) 0 ∞ indeterminate form of type 0 or ∞ ,then

f(x) f (x) lim = lim , x→a g(x) x→a g(x) provided the second limit exists.

68 l’Hˆopital’s Rule also applies to one-sided limits and limits where x →±∞. For example, this allows us to show that tan(3x) (tan(3x)) 3 sec2(3x) lim = lim  = lim =3 x→2 x  x→2 (x) x→2 1 0 Type 0

and 2x 2x 2x 2x e 2e 4e 8e ∞ lim 3 = lim 2 = lim = lim =+ . x→+∞ x  x→+∞3x  x→+∞ 6x x→+∞ 6 ∞ ∞ ∞ Type ∞ Type ∞ Type ∞ In this last example we had to apply l’Hˆopital’s rule multiple times. This can happen in practice, but it is important to make sure that the hypotheses of l’Hˆopital’s rule are satisfied each time before you apply it, for otherwise you will get horribly incorrect answers.

IMPORTANT. Note that l’Hˆopital’s Rule only applies directly to indeterminate forms of 0 ∞ type 0 or ∞ . The other types of indeterminate forms must first be reduced to one of these two types if we wish to apply l’Hˆopital’s Rule.

Example 82. To go back to our first example, we see that

e3x − 1 3e3x 3 · 1 lim = lim = =3. − − x→0 sin(x) x→0 cos(x) 1 0 Type 0

3x We can similarly compute that lim e −1 = 3, and therefore the graph of y = f(x)doesnot x→0+ sin(x) have a vertical asymptote at x =0.

Example 83 (Type 0 ·∞). The limit lim x2 ln(x) is an indeterminate form of type 0 ·∞. x→0+ 0 We use algebra to rewrite the function so that it becomes an indeterminate form of type 0 ∞ or (in this case) ∞ :

ln(x) 1 x2 lim x2 ln(x) = lim = lim x = lim − =0. + + 1 + 2 + x→0 x→0 2 x→0 − 3 x→0 2      x  x Type 0·∞ ∞ Type ∞ The indeterminate forms of type 00,1∞,and∞0 are all handled in the same way, so we’ll just do an example of 00 and ∞0.

2 Example 84 (Type 00). The limit lim xx is an indeterminate form of type 00.Aswe x→0+ did with logarithmic differentiation, we can reduce the exponentiation to multiplication by

69   x2 2 ln x 2 writing xx = e = ex ln(x). Notice that lim x2 ln(x) is an indeterminate form of type x→0+ 0 ·∞which we know how to solve! So we first compute

ln(x) x2 lim x2 ln(x) = lim = lim − =0. + + 1 + x→0 x→0 2 x→0 2   x  ∞ Type ∞

Because eu is a continuous function of u,weletu = x2 ln(x)toobtain

x2 x2 ln(x) lim x2 ln(x) 0 lim x = lim e = e x→0+ = e =1. x→0+ x→0+   eu is continuous

0 x 3 0 Example 85 (Type ∞ ). The limit lim (1 + 2e ) x is an indeterminate form of type ∞ . x→+∞  x 3 As before, we compute the limit of ln (1 + 2e ) x first, and then use continuity to compute x 3 the limit of (1 + 2e ) x :

  x x x x 3 3 ln(1 + 2e ) 6e 6e x lim ln (1+2e ) = lim = lim x = lim x =3. x→+∞ x→+∞  x  x→+∞ 1+2e x→+∞ 2e ∞ ∞ Type ∞ Type ∞

Therefore,

x x x 3 3ln(1+2e ) lim 3ln(1+2e ) 3 lim (1+2e ) x = lim e x = e x→+∞ x = e . x→+∞ x→+∞   eu is continuous Finally we see an example of type ∞−∞, which can often be solved by somehow combining the terms via algebra, properties of logarithms, etc.   1 1 Example 86 (Type ∞−∞). The limit lim − is an indeterminate form of type x→0+ sin(x) x ∞−∞. We use algebra to simplify it:   1 1 x − sin(x) 1 − cos(x) sin(x) 0 lim − = lim = lim = lim = =0. x→0+ x→0+ x→0+ x→0+ − − sin(x) x  xsin(x)   sin(x)+x cos(x) 2cos(x) x sin(x) 2 0 0 0 Type 0 Type 0

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