Indeterminate Forms and L’Hopital’s Rule Part 2

Chapter 4.4

April 17, 2007

Indeterminate Forms

Any limit of the form is called an indeterminate form of type .

We call the form an indeterminate form of type .

Other Indeterminate forms are:

limx→ a

f (x)g(x)

=000

0

limx→ a

f (x)g(x)

=∞∞

∞∞

0 ⋅∞ ,  ∞−∞ , 00  , ∞0  , 1∞

L’Hopital’s Rule applies ONLY to

indeterminate forms:

If the

Or

Then

Take for example:

limx→ a

f (x) =0 =limx→ a

g(x)

limx→ a

f(x) =∞ =limx→ a

g(x)

limx→ a

f(x)g(x)

=limx→ a

′f (x)′g (x)

limx→ 0

sin2 7x2x

⎛

⎝⎜⎞

⎠⎟0

0⎛⎝⎜

⎞⎠⎟

=limx→0

14sin 7x cos 7x

2=0

Indeterminate form:

Provided the limit on the right exists (or is or )∞ −∞

0

0⎛⎝⎜

⎞⎠⎟,

∞∞

⎛⎝⎜

⎞⎠⎟.

However we can manipulate expressions of other forms so they fit the criteria:

The form is:

L’Hopitals does not apply, but if we rewrite the limit as

the resulting form will be or

And we can apply L’Hopital’s Rule.

limx→ −∞

x2

1ex

0 ⋅∞ limx→ a

f(x)g(x)( ) =0⋅∞( )

limx→ a

f(x)1g(x)

 or limx→ a

g(x)1f(x)

0

0

∞∞

limx→ −∞

x2ex

= limx→−∞

x2

e− x= limx→−∞

2x

−e−x =0= limx→−∞

2

e−x

However we can manipulate expressions of other forms so they fit the criteria:

The form is:

L’Hopitals does not apply, but if we rewrite the limit as a quotient (finding a common denominator) the resulting

form will be or

And we can apply L’Hopital’s Rule.

limt→ 0+

t2 −sintt2 sint

∞−∞

0

0

∞∞

limt→ 0+

1sint

−1t2

⎛⎝⎜

⎞⎠⎟

= limt→0+

2t − cos t

t 2 cos t + 2t sin t =0 −1

0⎛⎝⎜

⎞⎠⎟ =−∞

0

0⎛⎝⎜

⎞⎠⎟

Example:

The form is:

L’Hopitals does not apply, but we can combine the log term using properties of logs:

Which gives us the form

limx→ ∞

lnx+ 5x

1x

∞⋅∞−∞( )

limx→ ∞

x ln x+ 5 −ln x( )

=5( )

limx→ ∞

x lnx+ 5x

⎛⎝⎜

⎞⎠⎟

∞⋅0

The last forms involve exponents and we’ll use these properties of logs:in our solution.

The form is:

L’Hopitals does not apply, but if we rewrite the function in terms of e using the first property of logs:

limx→ 0+

lnxx

00

limx→ 0+

elnxx

 

limx→ 0+

xx

= limx→0+

x ln x = limx→0+

ln x1

x

=0= limx→0+

1

x

−1

x2

00  , ∞0  , 1∞

eln x =x

lnxr =r lnx

=elimx→0+

ln xx

  =e?  

We can use the second property and find the limit using our previous forms.

∞∞

⎛⎝⎜

⎞⎠⎟

0 ⋅∞( )

= limx→0+

−x( )

e0   =1

More Examples:

limy→ 0+

1+ y( )1y

limx→ 0

cscx−cotx( )

limt→ 0+

t lnt( )

=e

=0

=0

Try:

limx→ 0

1+ sin 4x( )( )cotx

limx→ −∞

x2ex

More Examples:

limx→ 0

bx2 −5x+ sin 5x( )x3

⎛

⎝⎜⎞

⎠⎟Is finite.

Determine a value of b for which b= 0

Determine a value of b for which

limx→ 0

bx2 + 4x+ log 1−4x( )x3

⎛

⎝⎜⎞

⎠⎟Is finite. b= 8

Determine the behavior of the limits:

limx→ 0

1xlogx

⎛⎝⎜

⎞⎠⎟

limx→ ∞

xlogx

⎛⎝⎜

⎞⎠⎟

Both diverge