∞ / ∞, 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 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 evaluate 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:__

Note that

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

__Exercises:__ Find the limits

1. lim_{x→∞} (ln x)^{ 1/x}

2. lim_{x→∞} [ ln x - ln (1 + x) ]

3. lim_{x→∞} x / e^{ x}

4. lim_{x→0+} x ^{ sin x}

__Solutions to Above Exercises:__ Find the limits

1. 1

2. 0

3. 0

4. 1

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