## 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.

**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**

__Theorem:__lim [ f(x) / g(x) ] = lim [ f'(x) / g'(x) ]

lim stands for lim

_{x→a}, lim

_{x→a+}, lim

_{x→a-}, lim

_{x→ + ∞}or lim

_{x→ - ∞}.

__Example 1:__ Find the limit lim_{x→∞} ln x / x

__Solution to Example 1:__

Since

lim

_{x→∞}ln x = ∞

and

lim

_{x→∞}x = ∞

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

lim

_{x→∞}ln x / x = lim

_{x→∞}[ d ( ln x ) / dx ] / [ d ( x ) / dx ]

= lim

_{x→∞}[ 1 / x ] / 1 = 0

__Example 2:__ Find lim_{x→∞} x e^{ -x}

__Solution to Example 2:__

lim

_{x→∞}x = + ∞

and

lim

_{x→∞}e

^{ - x}= 0

This is the indeterminate form ∞

^{ . }0. The idea is to convert it into to the indeterminate form ∞ / ∞

lim

_{x→∞}x e

^{ -x}= lim

_{x→∞}x / e

^{ x}

Now apply the above L'hopital's theorem

lim

_{x→+∞}x / e

^{ x}= lim

_{x→+∞}1 / e

^{ x}= 0

__Example 3:__ Find lim_{x→∞} ( 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

lim_{x→+∞} ( 1 + 1/x)^{ x} = lim_{t→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

lim_{t→0} (1 / t) ln (1 + t) = lim_{t→ 0} [ d ( ln(1 + t) ) / dt ] / [ d ( t ) / dt ]

= lim_{t→0} [ 1 / (1 + t) / 1 = 1

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

lim_{x→∞} ( 1 + 1 / x)^{ x} = e

__Example 4:__ Find the limit lim_{x→0+} (1 / x - 1 / sin x)

__Solution to Example 4:__

Note that

lim_{x→0+} 1 / x = +∞

and

lim_{x→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

lim_{x→0+} (1 / x - 1 / sin x) = lim_{x→0+} [ (sin x - x) / (x sin x) ]

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

lim_{x→0+} [ (sin x - x) / (x sin x) ]

= lim_{x→0+} [ (cos x - 1) / (sin x + x cos x) ]

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

lim_{x→0+} [ (sin x - x) / (x sin x) ]

= lim_{x→0+} [ (cos x - 1) / (sin x + x cos x) ]

= lim_{x→0+} [ (-sin x) / (cos x + cos x - x sin x) ] = 0 / 2 = 0

__Example 5:__ Find the limit lim_{x→ 0+} x^{ x}

__Solution to Example 5:__

We have the indeterminate form 0^{0}. Let y = x^{ x} and ln y = ln (x^{ x}) = x ln x. Let us now find the limit of ln y

lim_{x→0+} ln y

= lim_{x→0+} x ln x

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

lim_{x→0+} x ln x

= lim_{x→0+} ln x / (1 / x)

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

lim_{x→0+} ln x / (1 / x)

= lim_{x→0+} (1 / x) / (- 1 / x^{ 2})

= lim_{x→0+} -x = 0

The limit of ln y = 0 and the limit of y = x^{x} 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}, lim

_{x→a+}, lim

_{x→a-}, lim

_{x→+ ∞}or lim

_{x→ - ∞}.

__Example 1:__ Find the limit lim_{x→0} sin x / x

__Solution to Example 1:__

Since

lim_{x→0} sin x = 0

and

lim

_{x→0}x = 0

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

lim

_{x→0}sin x / x = lim

_{x→0}[ d ( sin x ) / dx ] / [ d ( x ) / dx ]

= lim

_{x→0}cos x / 1 = 1

__Example 2:__ Find the limit lim_{x→0} ( e^{ x} - 1 ) / x

__Solution to Example 2:__

Note that

lim_{x→0} ( e^{ x} - 1 ) = 0

and

lim_{x→0} x = 0

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

lim_{x→0} ( e^{ x} - 1 ) / x = lim_{x→0} [ d ( e^{ x} - 1 ) / dx ] / [ d ( x ) / dx ]

= lim_{x→0} e^{ x} / 1 = 1

__Example 3:__ Find the limit lim_{x→1} ( x^{ 2} - 1 ) / (x - 1)

__Solution to Example 3:__

Since the limit of the numerator

lim

_{x→1}( x

^{ 2}- 1 ) = 0

and that of the denominator

lim

_{x→1}x - 1 = 0

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

lim

_{x→1}( x

^{ 2}- 1 ) / (x - 1) = lim

_{x→1}[ d ( x

^{ 2}- 1 ) / dx ] / [ d ( x - 1 ) / dx ]

= lim

_{x→1}2 x / 1 = 2

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

lim

_{x→1}( x

^{ 2}- 1 ) / (x - 1)

= lim

_{x→1}( x - 1 )(x + 1) / (x - 1) =

= lim

_{x→1}(x + 1) / 1 = 2

__Example 4:__ Find the limit lim_{x→2} ln(x - 1 ) / (x - 2)

__Solution to Example 4:__

Limit of numerator

lim_{x→2} ln(x - 1) = 0

Limit of denominator

lim_{x→2} x - 1 = 0

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

lim_{x→2} ln(x - 1) / (x - 2) = lim_{x→2} [ d ( ln(x - 1) ) / dx ] / [ d ( x - 2 ) / dx ]

= lim_{x→2} [ 1 / (x-1) ] / 1 = 1

__Example 5:__ Find the limit lim_{x→0} (1 - cos x ) / 6 x^{ 2}

__Solution to Example 5:__

Limit of numerator and denominator

lim_{x→0} 1 - cos x = 0

lim_{x→0} 6 x^{ 2} = 0

L'hopital's rule may be used

lim_{x→0} (1 - cos x ) / 6 x^{ 2} = lim_{x→0} [ sin x ] / [ 12 x ]

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

lim_{x→0} (1 - cos x ) / 6 x^{ 2} = lim_{x→0} [ sin x ] / [ 12 x ]

= lim_{x→0} cos x / 12 = 1 / 12

__Exercises:__ Find the limits

1. lim_{x→0} (sin 4x / sin 2x)

2. lim_{x→0} tan x / x

3. lim_{x→1} ln x / (3x - 3)

4. lim_{x→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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