Indeterminate Forms -L’Hôpital’s Rule

correct spellings: “L’Hôpital” and “L’Hospital”) Sometimes, we run into indeterminate forms. These are things like
00
and
??
Forinstance,howdoyoudealwiththefollowing?
lim x3 ? 1 = 0??
x1 x2 ? 10
?
Example 0. Onewayofdealingwiththisistousealgebratosimplifythings:
lim x3 ? 1 = lim (x ? 1)(x2 + x + 1) = lim x2 + x +1 = 3 x?1 x2 ? 1 x?1 (x ? 1)(x + 1) x?1 x +1 2
Ingeneral,when f(a)= g(a)=0, f(x) f(x) xlimaf(x) ? f(a) f?(a)lim = lim x ? a = ?x ? a =
x?a g(x) x?a g(x) lim g(x) ? g(a) g?(a) xa
x ? a ?x ? a
ThisistheeasyversionofL’Hôpital’srule:
f(x) f?(a)
lim =
x?a g(x) g?(a)
Note: this only works when g?(a)=0?! Inexample0,
f(x)= x3 = 1; g(x)= x2 ? 1
f?(x)=3x2; g?(x)=2x = f?(1) = 3; g?(1) = 2
?
Thelimitis f?(1)/g?(1) = 3/2. Now, let’s go on to the full L’Hôpital rule.
1
???
???
? ?
Lecture 34 18.01 Fall 2006
Example 1. Apply L’Hôpital’s rule (a.k.a. “L’Hop”) to
x
lim 15 ? 1
x1 x3
?? 1
toget
lim x15 ? 1 = lim 15x14 = 15 =5
x?1 x3 ? 1 x?1 3x2 3
Let’scomparethiswiththeanswerwe’dgetifweusedlinearapproximationtechniques,insteadof L’Hôpital’srule:
x15 ? 1 ? 15(x ? 1)
(Here, f(x)= x15 ? 1,a =1,f(a)= b =0,m = f?(1) = 15,and f(x) ? m(x ? a)+ b.) Similarly,
x3 ? 1 ? 3(x ? 1)
Therefore,
x15 ? 1 15(x ? 1) =5 x3 ? 1 ? 3(x ? 1)
Example 2. ApplyL’Hopto
sin3x
lim
x0 x
?
toget
3cos3x
lim =3
x0 1
?
Thisisthesameas
d
sin(3x) = 3 cos(3x) =3
dx
x=0 x=0
Example 3.
lim sin x ? cos x = lim cos x + sin x =1+1= ?2
? ? ?
44
x? x ? 4 x? 1 ?2 ?2
f(x) = sin x ? cos x, f?(x) = cos x + sin x = ?2
f? ?
4
?y 0
Remark: Derivatives lim are always a type of limit.
?x0 ?x 0
?
Example 4. lim cos x ? 1.
x0 x
?
UseL’Hôpital’sruletoevaluatethelimit:
lim cos x ? 1 = lim ?sin x =0
x0 x x0 x
??
2
Lecture 34 18.01 Fall 2006
Example 5. lim cosxx2? 1.
x0
?
cos x ? 1 cos x ? 1 ?sin x ?cos x 1
xlim0 x2 = xlim0 x2 = xlim0