Lhospitals Rule For Indeterminate Forms Homework Sheets

Examples with detailed solutions and exrcises that solves limits questions related to inderminate forms such as

∞ / ∞, 0 0, ∞ 0, 1, ∞ o and ∞ - ∞.

A second version of L'hopital's rule allows us to replace the limit problem ∞ / ∞ with another simpler problem to solve.

Theorem: If lim f(x) = ∞ and lim g(x) = ∞ and if lim [ f'(x) / g'(x) ] has a finite value L , or is of the form + ∞ or - ∞, then

lim [ f(x) / g(x) ] = lim [ f'(x) / g'(x) ]

lim stands for limx→a, limx→a+, limx→a-, limx→ + ∞ or limx→ - ∞.

Example 1: Find the limit limx→∞ ln x / x

Solution to Example 1:



Since

limx→∞ ln x = ∞

and

limx→∞ x = ∞

The above L'hopital's rule can be used to evalute the given limit question

limx→∞ ln x / x = limx→∞ [ d ( ln x ) / dx ] / [ d ( x ) / dx ]

= limx→∞ [ 1 / x ] / 1 = 0

Example 2: Find limx→∞ x e -x

Solution to Example 2:

Note that

limx→∞ x = + ∞

and

limx→∞ e - x = 0

This is the indeterminate form ∞ . 0. The idea is to convert it into to the indeterminate form ∞ / ∞

limx→∞ x e -x = limx→∞ x / e x

Now apply the above L'hopital's theorem

limx→+∞ x / e x = limx→+∞ 1 / e x = 0

Example 3: Find limx→∞ ( 1 + 1/x) x

Solution to Example 3:

The above limit is of the indeterminate form 1. If we let t = 1 / x the above limit may written as

limx→+∞ ( 1 + 1/x) x = limt→0 ( 1 + t) 1 / t

Let y = ( 1 + t) 1 / t and find the limit of ln y as t approaches 0

ln y = (1 / t) ln (1 + t)

The advantage of using ln y is that now the limit has the indeterminate form 0 / 0 and the first L'hopital's rule can be applied

limt→0 (1 / t) ln (1 + t) = limt→ 0 [ d ( ln(1 + t) ) / dt ] / [ d ( t ) / dt ]

= limt→0 [ 1 / (1 + t) / 1 = 1

Since the limit of ln y = 1 the limit of y is e 1 = e, hence

limx→∞ ( 1 + 1 / x) x = e


Example 4: Find the limit limx→0+ (1 / x - 1 / sin x)

Solution to Example 4:

Note that

limx→0+ 1 / x = +∞

and

limx→0+ 1 / sin x = +∞

This limit has the indeterminate form ∞ - ∞ and has to be converted to another form by combining 1 / x - 1 / sin x

limx→0+ (1 / x - 1 / sin x) = limx→0+ [ (sin x - x) / (x sin x) ]

We now have the indeterminate form 0 / 0 and we can use the L'hopital's theorem.

limx→0+ [ (sin x - x) / (x sin x) ]

= limx→0+ [ (cos x - 1) / (sin x + x cos x) ]

We have the indeterminate form 0 / 0 and use the L'hopital's theorem again.

limx→0+ [ (sin x - x) / (x sin x) ]

= limx→0+ [ (cos x - 1) / (sin x + x cos x) ]

= limx→0+ [ (-sin x) / (cos x + cos x - x sin x) ] = 0 / 2 = 0

Example 5: Find the limit limx→ 0+ x x

Solution to Example 5:

We have the indeterminate form 00. Let y = x x and ln y = ln (x x) = x ln x. Let us now find the limit of ln y

limx→0+ ln y

= limx→0+ x ln x

The above limit has the indeterminate form 0 . ∞. We have convert it as follows

limx→0+ x ln x

= limx→0+ ln x / (1 / x)

It now has the indeterminate form ∞ / ∞ and we can use the L'hopital's theorem

limx→0+ ln x / (1 / x)

= limx→0+ (1 / x) / (- 1 / x 2)

= limx→0+ -x = 0

The limit of ln y = 0 and the limit of y = xx is equal to

e 0 = 1

L'hopital's rule allows us to replace a limit problem with another that may be simpler to solve.

Theorem: If lim f(x) = 0 and lim g(x) = 0 and if lim [ f'(x) / g'(x) ] has a finite value L , or is of the form + ∞ or - ∞, then

lim [ f(x) / g(x) ] = lim [ f'(x) / g'(x) ]

lim stands for lim

x→a, limx→a+, limx→a-, limx→+ ∞ or limx→ - ∞.

Example 1: Find the limit limx→0 sin x / x

Solution to Example 1:

Since

limx→0 sin x = 0

and

lim

x→0 x = 0

L'hopital's rule can be used to evalute the above limit as follows

limx→0 sin x / x = limx→0 [ d ( sin x ) / dx ] / [ d ( x ) / dx ]

= limx→0 cos x / 1 = 1

Example 2: Find the limit limx→0 ( e x - 1 ) / x

Solution to Example 2:

Note that

limx→0 ( e x - 1 ) = 0

and

limx→0 x = 0

We can use L'hopital's rule to calculate the given limit as follows

limx→0 ( e x - 1 ) / x = limx→0 [ d ( e x - 1 ) / dx ] / [ d ( x ) / dx ]

= limx→0 e x / 1 = 1

Example 3: Find the limit limx→1 ( x 2 - 1 ) / (x - 1)

Solution to Example 3:



Since the limit of the numerator

limx→1 ( x 2 - 1 ) = 0

and that of the denominator

limx→1 x - 1 = 0

are both equal to zero, we can use L'hopital's rule to calculate limit

limx→1 ( x 2 - 1 ) / (x - 1) = limx→1 [ d ( x 2 - 1 ) / dx ] / [ d ( x - 1 ) / dx ]

= limx→1 2 x / 1 = 2

Note that the same limit may be calculated by first factoring as follows

limx→1 ( x 2 - 1 ) / (x - 1)

= limx→1 ( x - 1 )(x + 1) / (x - 1) =

= limx→1 (x + 1) / 1 = 2

Example 4: Find the limit limx→2 ln(x - 1 ) / (x - 2)

Solution to Example 4:

Limit of numerator

limx→2 ln(x - 1) = 0

Limit of denominator

limx→2 x - 1 = 0

Both limits are equal to zero, L'hopital's rule may be used

limx→2 ln(x - 1) / (x - 2) = limx→2 [ d ( ln(x - 1) ) / dx ] / [ d ( x - 2 ) / dx ]

= limx→2 [ 1 / (x-1) ] / 1 = 1

Example 5: Find the limit limx→0 (1 - cos x ) / 6 x 2

Solution to Example 5:

Limit of numerator and denominator

limx→0 1 - cos x = 0

limx→0 6 x 2 = 0

L'hopital's rule may be used

limx→0 (1 - cos x ) / 6 x 2 = limx→0 [ sin x ] / [ 12 x ]

The new limit is also indeterminate 0/0 and we may apply l'hopital's theorem a second time

limx→0 (1 - cos x ) / 6 x 2 = limx→0 [ sin x ] / [ 12 x ]

= limx→0 cos x / 12 = 1 / 12



Exercises: Find the limits

1. limx→0 (sin 4x / sin 2x)

2. limx→0 tan x / x

3. limx→1 ln x / (3x - 3)

4. limx→0 ( e x - 1 ) / sin 2 x

Solutions to Above Exercises:

1. 2

2. 1

3. 1 / 3

4. 1 / 2


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