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l’Hospital’s Rule ~ page 1
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
OBJECTIVE: Use l’Hospital’s Rule to find limits of indeterminate forms
l’Hospital’s Rule is a handy shortcut for finding the limit of some functions l(x) which are of the form
. We first determine if the given function meets the prerequisites for l’Hospital’s Rule.
f(x) l(x) g(x)
- Necessary conditions for applying l’Hospital’s Rule
- f(x) and g(x) must both be differentiable functions near a (not necessarily at a )
- g (x)′^ ≠^0 near a (except possibly at a)
- these are called indeterminate forms x a
f(x) 0 lim or ; → g(x) 0
∞
∞ (^) x
f(x) 0 lim or ; →±∞g(x) 0
- IF l’Hospital’s Rule applies, then the result of applying it is:
- Note the use of ; this is not an equality in the normal sense of
H
x a x a
f(x) f (x) lim lim → g(x) → g (x)
H
=
the word.
- Always simplify after applying l’Hospital’s Rule. If the simplified result meets all the
necessary conditions , then l’Hospital’s Rule may be applied again. Always simplify
before applying the rule again!
- Using l’Hospital’s Rule when it doesn’t apply is wrong , even though it may give the “right
answer”.
Example 1: Find
2
x^2
4x 3x 6 lim → ∞ 5 2x
- We already know one way to find this limit!
= =
2
x^2
4x 3x 6 lim → ∞ 5 2x
2
2 2 2
x^2
2 2
4x 3x 6
x x x lim 5 2x
x x
→∞
− +
−
2
x 2
0 0
3 6 4 x (^) x lim 5 2 x
0
→∞
− +
−
/ /
2
- Check to see if l’Hospital’s Rule applies:
- The numerator and denominator of the function are both polynomials, which are
differentiable everywhere
- The derivative of the denominator is not equal to zero as x increases without bound
- = , which is one of the indeterminant forms
2
x^2
4x 3x 6 lim → ∞ 5 2x
− +
−
- Therefore, l’Hospital’s Rule applies!
l’Hospital’s Rule ~ page 2
2
x^2
4x 3x 6 lim → ∞ 5 2x
H
= x
8x 3 lim → ∞ 4x
- We can show that the new function meets all the conditions of l’Hospital’s
8x 3 m(x) 4x
Rule and that =. x
8x 3 lim → ∞ 4x
∞
∞
2
x^2
4x 3x 6 lim → ∞ 5 2x
H
= x
8x 3 lim → ∞ 4x
H
= x
lim 2 → ∞ 4
The previous example is not very impressive because we already knew how to find the derivative.
However, l’Hospital’s Rule will enable us to find other types of limits.
Example 2: Find x 0
lim → 2
sin x
x +x
- Check to see if l’Hospital’s Rule applies:
- The numerator and denominator are both differentiable functions near 0 (and actually at 0,
in this case)
- The derivative of the denominator is 0 at x = 0, but it is not zero in the neighborhood of 0
- = , which is one of the indeterminate forms x 0
lim → 2
sin x
x +x
- Therefore, l’Hospital’s Rule applies
- Thus, = x 0
lim → 2
sin x
x +x
H
= x 0
lim →
co s x
1 +2x
Example 3: Find
x
lim → ∞
2x
3
e
x
2
- It can be shown that l’Hospital’s Rule applies to this limit, and
- = x
lim → ∞
2x
3
e
x
2 ∞
lim → ∞
2x
3
e
x
2 H
= x
lim → ∞
2x^2
2
4xe
3x x
lim → ∞
2x^2 4e
3x
- l’Hospital’s Rule applies again and = x
lim → ∞
2 2x 4e
3x
lim → ∞
2x
3
e
x
2 H
= x
lim → ∞
2 2x
2
4xe
3x x
lim → ∞
2x^2 4e
3x
H
= x
lim → ∞
2x^2 16xe
∞
l’Hospital’s Rule ~ page 4
Find
2
x
3 x lim x e →∞
−
Example 6: Find
x 0
lim →
(csc x −cot x)
- = This is an indeterminate difference of type : x 0
lim →
(csc x − cot x) ∞ − ∞ ∞ − ∞
lim →
(csc x − cot x) x 0
lim →
1 cos x
sin x sin x
⎝ ⎠ x^0
lim →
1 cos x
sin x
- Now we can see that l’Hospital’s Rule applies and
- = x 0
lim →
1 cos x
sin x
lim →
(csc x − cot x) x 0
lim →
1 cos x
sin x
H
= x 0
lim →
sin x
cos x x^0
lim →
tan x
Find x 1
lim → ln x^ x^1
Indeterminate power types include. They are discussed on page 302.
0 0 0 , , and 1
∞ ∞
l’Hospital’s Rule ~ page 5
Solutions
x 0
x tan x lim → sin x
H
x 0 x 0
2 x tan x 1 sec x 1 1 lim lim 2 → sin x → cos x 1
x 0
x x 5 3 0 lim → x^0
H
x 0 x 0
x x x x 5 3 5 ln 5 3 ln 3 5 lim lim ln 5 ln 3 ln → x^ →^1
x 0 2
cosmx cosnx lim → x
H
x 0 2 x 0
cosmx cosnx m sinmx n sinnx 0 lim lim → (^) x → 2x^0
H H
x 0 x 0 x 0
2 2
2
cosmx cosnx m sinmx n sinnx m cosmx n cosnx lim lim lim → (^) x → 2x^ →^2
2 2 m n (^1 2 ) (n m ) 2 2
2
x
3 x lim x e →∞
−
x^2
3
x
x lim
e
→∞
2 2 2
H
x x x
3 2
x x x
x x x lim lim lim
e 2xe 2e
→∞ →∞ →∞
2 2
H
x x^ x x
x 1 lim lim 0
2e 4xe
→∞ →∞
x 1
lim → ln x^ x^1
⎝ − ⎠ x^1
x 1 ln x 0 lim → ln x(x^ 1)^0
⎛ −^ − ⎞
H
x 1 x 1 x 1
x 1 ln x (^) x x x 1 0 lim lim lim ln x(x 1) 1 x x ln x x 1 0 (ln x)(1) (x 1) x
→ → →
⎝ −^ ⎠ ⎝ +^ − ⎠
i
H
x 1 x 1 x 1
x 1 1 1 1 lim lim lim x ln x x 1 1 ln x 2 2 x ln x 1 x
→ → →
= ⎜^ ⎟= =
⎝ +^ −^ ⎠ ⎜^ ⎛^ ⎞ ⎟ +
⎜ ⎟+^ +
⎝ ⎝^ ⎠ ⎠
