Intermediate Forms and L' Hospital's Rule

ü David A. Smith

Notes

We say that the following limit

limxض3x-1

x3

has the intermediate form ¶¶

because 3x ض and x3 ض as x ض. We will consider the following indeterminate

forms,

¶

¶

0

000 ¶0 ¶-¶ 0 ÿ¶ 1¶

Our main investigative tool will be L'Hospital's rule which says that the limit of a quotient of functions f and g is equalto the limit of the quotient of their derivatives f ' and g ', provided that the following conditions are satisfied. It isespecially important to verify the conditions regarding the limits of f and g before applying L'Hospital's rule. Also,keep in mind that L'Hospital's' rule also holds if "x Ø c" is replaced by x Ø c+, x Ø c-, x ض, x Ø-¶.

Proposition (L'Hospital's Rule) Suppose f and g are differentiable functions and g ' HxL 0 on an open interval I that

contains c (except possibly at c). Suppose limx-cf HxLgHxL produces an intermediate form 0

0 or ¶

¶ and that limxØc

f ' HxLg' HxL = L,

then

limxØcf HxLgHxL = limxØc

f ' HxLg' HxL .

Example (L'Hospital's Rule with Intermediate Form ¶¶

) Evaluate the limit limxضln x

x using L'Hospital's rule.

Solution. The limit limxØ+¶ln x

x has the indeterminate form ¶

¶ since limxض ln x = +¶ and limxØ+¶ x = +¶. We

try to use L'Hospital's rule to find,

limxØ+¶ln x

x= limxØ+¶

1

x

1= 0.

É

Example (L'Hospital's Rule with Intermediate Form ¶¶

) Evaluate the limit limxØ-¶x2

‰-x using L'Hospital's rule.

Solution. The limit limxØ-¶x2

‰-x has the indeterminate form ¶

¶ since limxØ-¶ x2 = +¶ and limxØ-¶‰-x = +¶.

We try to use L'Hospital's rule to find,

limxØ-¶x2

‰-x= limxØ-¶

2 x

-‰-x= limxØ-¶

2

‰-x= 0

Thus illustrating that L'Hopital's rule can be used multiple times. É

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Example (L'Hospital's Rule with Intermediate Form 00) Evaluate the limit limxØ0+ Hsin xLx using L'Hospital's rule.

Solution. The limit limxØ0+ Hsin xLx has the indeterminate form 00 since limxØ0+= sin x and limxØ0+ x = 0.

Suppose the limit exists, say limxØ0+ Hsin xL1êx = L. Then,

ln L = ln @limxØ0+ Hsin xLxD

ln L = limxØ0+ ln Hsin xLx

ln L = limxØ0+ln Hsin xL

1

x

Since limxØ0+ ln Hsin xL = +¶ and limxØ0+1

x= +¶ this limit has the indeterminate form of ¶

¶ and so we try L'Hospital's

rule,

ln L = limxØ0+

cos x

sin x

-1

x2

ln L = limxØ0+-x2

tan x

ln L = limxØ0+-2 x

sec2 x

ln L = 0

L = ‰0 = 1

É

Example (L'Hospital's Rule with Intermediate Form ¶0) Evaluate the limit limxØ+¶ x1êx using L'Hospital's rule.

Solution. The limit limxØ+¶ x1êx has the indeterminate form ¶0 since limxض1

x= 0 and limxØ+¶ x = +¶.

Suppose the limit exists, say limxØ+¶ x1êx = L. Then,

ln L = ln limxØ+¶ x1êx

ln L = limxØ+¶ Iln x1êxM

ln L = limxØ+¶ln x

x

ln L = limxØ+¶

1

x

1

ln L = 0

L = ‰0 = 1

É

2

Example (L'Hospital's Rule with Intermediate Form 1¶) Evaluate the limit limxØ0+ H1 + sin 4 xLcot x using L'Hospital's

rule.

Solution. The limit limxØ0+ H1 + sin 4 xLcot x has the indeterminate form 1¶ since limxØ0+ 1 + sin 4 x = 1 and

limxØ0+ cot x = +¶. Suppose the limit exists, say limxØ0+ H1 + sin 4 xLcot x = L. Then,

ln L = ln limxØ0+ H1 + sin 4 xLcot x

ln L = limxØ0+ ln@H1 + sin 4 xLcot xD

ln L = limxØ0+ @cot x lnH1 + sin 4 xLD

ln L = limxØ0+lnH1+sin 4 xL

tan x

The limit limxØ0+lnH1+sin 4 xL

tan x has the indeterminate form 0

0 since limxØ0+ lnH1 + sin 4 xL = 0 and limxØ0+ tan x = 0. So we

try to use L'Hospital's rule to find,

ln L = limxØ0+

4 cos 4 x

1+sin 4 x

sec2 x= 4

ln L =

4

1+0

1= 4

L = ‰4.

É

Example (L'Hospital's with Rule Intermediate Form 0 ÿ¶) Evaluate the limit limxØ+¶‰-x x using L'Hospital's rule.

Solution. The limit limxØ+¶‰-x x has the indeterminate form 0 ÿ¶ since limxØ+¶‰-x = 0 and

limxØ+¶ x = +¶. We try L'Hospital's rule to find,

limxØ+¶x

‰x= limxØ+¶

1

2 x

‰x= limxØ+¶

1

2 x ‰x= 0.

É

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Example (L'Hospital's Rule with Intermediate Form ¶ -¶) Evaluate the limit limxØ+¶ x5BsinJ 1xN - 1

x+

1

6 x3F using

L'Hospital's rule.

Solution. Let's make a change of variable to simply the express, namely u =1

x. Since u Ø 0 as x Ø+¶, we have

limxØ+¶ x5BsinJ 1xN - 1

x+

1

6 x3F

= limuØ0

sinHuL-u+ 1

6u3

u5

= limuØ0

sinHuL-u+ 1

6u3

u5

The limit limuØ0

sinHuL-u+ 1

6u3

u5 has an indeterminate form 0

0 and so we try L'Hopital's rule,

= limuØ0Hcos uL-1+ 1

2u2

5 u4

= limuØ0H-sin u L+u20 u3

= limuØ0Hcos u L+120 u3

= limuØ0H-sin u L60 u2

= limuØ0H-cos u L120 u

=1

120

É

Example (Using L'Hospital's Rule Incorrectly) Try to evaluate the limit limxØ+¶x+sin x

x-cos x using L'Hospital's rule.

Solution. This limit has indeterminate form since

limxØ+¶ Hx + sin xL = +¶

4

and

limxØ+¶ Hx + cos xL = +¶.

If we try to apply L'Hospitals's Rule we find,

limxØ+¶x+sin x

x-cos x= limxض

1+cos x

1+sin x

but the limit limxض1+cos x

1+sin x does not exist because of osculating behavior, so we can not use L'Hospital's rule. To

correctly find this limit we divide by x, to find

limxØ+¶x+sin x

x-cos x= limxØ+¶

x

x+sin x

x

x

x-cos x

x

=1+0

1-0= 1.

É

Example (L'Hospital's Rule with Intermediate Form 00) Evaluate the limit limxØ0

‰2 x-1

x using L'Hospital's rule.

Solution. The limit limxØ0‰2 x-1

xhas the indeterminate form of 0

0 since limxØ0‰

2 x - 1 = 0 and limxØ0 x = 0. We

try to apply L'Hospital's rule to find,

limxØ0‰2 x-1

x= limxØ0

2 ‰2 x

1= 2.

É

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Exercises

(1) Use L'Hospital's rule to find the limit limxØ2x-2

x2-4.

(2) Use L'Hospital's rule to find the limit limxØ1x3-1

4 x3-x-3.

(3) Use L'Hospital's rule to find the limit limxØ-5x2-25

x+5.

(4) Use L'Hospital's rule to find the limit limxØ0sin x2

x.

(5) Use L'Hospital's rule to find the limit limxØ0sin x-x

x3.

(6) Use L'Hospital's rule to find the limit limxØpê21-sin x

1+ cos 2 x.

(7) Use L'Hospital's rule to find the limit limxØpê2ln H csc xLJx- p

2N2

.

(8) Use L'Hospital's rule to find the limit limxØJ p

2N- Ix -

p

2M sec x.

(9) Use L'Hospital's rule to find the limit limxØ0

J 12Nx-1

x.

(10) Use L'Hospital's rule to find the limit limxØ0+ln H‰x-1Lln x

.

(11) Use L'Hospital's rule to find the limit limxض Hln 2 x - lnHx + 1LL.

(12) Use L'Hospital's rule to find the limit limxØ0+ J 3 x+1x -1

sin xN.

(13) Use L'Hospital's rule to find the limit limxØ0cos x-1

‰x-x-1.

(14) Use L'Hospital's rule to find the limit limxض x2 ‰-x.

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(15) Use L'Hospital's rule to find the limit limxض Hln xL1êx.

(16) Use L'Hospital's rule to find the limit limxض x1ê ln x.

(17) Use L'Hospital's rule to find the limit limxض H1 + 2 xL1ê 2 ln x.

(18) Use L'Hospital's rule to find the limit limxØ0+x

sin x.

(19) Let f HxL = : x + 2 x 0

0 x = 0> and gHxL = : x + 1 x 0

0 x = 0>. Show that limxØ0

f ' HxLg' HxL = 1 but limxØ0

f HxLgHxL = 2. Explain why

this does not contradict L'Hospital's Rule.

(20) Use L'Hospital's rule to find constant a and b so that

limxØ0 J sin 2 xx3

+a

x2+bN = 1.

(21) Use L'Hospital's rule to find all values of a and b so that

limxØ0sin a x+b x

x3= 36.

(22) Use L'Hospital's rule to find the values of a so that

limxØ0a-cos b x

x2= 2.

(23) For a certain value of a, the limit limxØ+¶ Ix4 + 5 x3 + 3Ma - x is finite and nonzero. Find a and then use L'Hospital's

rule to compute the limit.

(24) Use L'Hospital's rule to evaluate the limit

limxØa2 a3 x-x4 -a a2 x

3

a- a x34

.

(25) Use L'Hospital's rule to determine which values of constants a and b is it true that

limxØ0 Ix-3 sin 7 x + a x-2 + bM = -2?

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(26) Use L'Hospital's rule to find constant a and b so that

limxØ0 J sin 2 xx3

+a

x2+bN = 1.

(27) Use L'Hospital's rule to find all values of a and b so that

limxØ0sin a x+b x

x3= 36.

(28) Use L'Hospital's rule to find the values of a so that

limxØ0a-cos b x

x2= 2.

(29) For a certain value of a, the limit limxØ+¶ Ix4 + 5 x3 + 3Ma - x is finite and zero. Use L'Hospital's rule to find a and

then compute the limit.

(30) Use L'Hospital's rule to evaluate the limit

limxØa2 a3 x-x4 -a a2 x

3

a- a x34

.

(31) Use L'Hospital's rule to determine which values of constants a and b is it true that

limxØ0 Ix-3 sin 7 x + a x-2 + bM = -2?

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