12 l'hopital rule ii

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

Theorem: Let k > 0, then lim k1/x = 1.x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1K

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1K K1/2

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1K K1/2 K1/3

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1K K1/2 K1/3 K1/4

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK K1/2 K1/3 K1/4

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK1/2K K1/2 K1/3 K1/4

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4

But instead of having a constant base k, what willhappen if k changes?

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4

But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a?

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4

But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a? We say these limit problems are of the form ∞0.

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4

But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a? We say these limit problems are of the form ∞0.

Similarly, if b(x) 0 and e(x) 0 as x a, we say these problems "lim b(x)e(x)" are of the form 00.

xa

Theorem: Let k > 0, then lim k1/x = 1.

In picture:x∞

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4

But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a? We say these limit problems are of the form ∞0.

Similarly, if b(x) 0 and e(x) 0 as x a, we say these problems "lim b(x)e(x)" are of the form 00.

xa

Lastly, we have the 1∞ forms, i.e. the problems of "lim b(x)e(x)" where b(x) 1 and e(x) ∞.

xa

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) . (∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. x∞

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

We find the limit of Ln(x1/x) = first.Ln(x)x

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

this is ∞/∞ form so use the L'Hopital's Rule.

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

=

this is ∞/∞ form so use the L'Hopital's Rule.

lim [Ln(x)]'[x]'x∞

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

=

this is ∞/∞ form so use the L'Hopital's Rule.

lim [Ln(x)]'[x]'x∞

=

lim 1xx∞

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

=

this is ∞/∞ form so use the L'Hopital's Rule.

lim [Ln(x)]'[x]'x∞

=

lim 1xx∞

= 0.

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

=

this is ∞/∞ form so use the L'Hopital's Rule.

lim [Ln(x)]'[x]'x∞

=

lim 1xx∞

= 0.

So lim x1/x =x∞

lim eLn(x )1/x

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

=

this is ∞/∞ form so use the L'Hopital's Rule.

lim [Ln(x)]'[x]'x∞

=

lim 1xx∞

= 0.

So lim x1/x =x∞

lim eLn(x )1/x

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

=

this is ∞/∞ form so use the L'Hopital's Rule.

lim [Ln(x)]'[x]'x∞

=

lim 1xx∞

= 0.

So lim x1/x =x∞

lim eLn(x )1/x

(∞0, 00, 1∞)

0

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

These are the three indeterminate forms related tolimits of power functions b(x)e(x) .

We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).

Example: A. Find lim x1/x. ( ∞0 form.)x∞

lim Ln(x1/x) =

We find the limit of Ln(x1/x) = first.Ln(x)x

lim Ln(x)xx∞ x∞

=

this is ∞/∞ form so use the L'Hopital's Rule.

lim [Ln(x)]'[x]'x∞

=

lim 1xx∞

= 0.

So lim x1/x =x∞

lim eLn(x )1/x = e0 = 1 .

0

(∞0, 00, 1∞)

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+ x0+

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+ x0+

lim Ln(x) x0+ x-1 =

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+ x0+

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+ x0+

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2

–x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+ x0+

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 0

–x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x

x0+

x0+

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 0

–x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x

x0+

x0+

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

–x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

–x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

–x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] lim x∞

lim x∞

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

–x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] lim x∞

lim x∞

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1=

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= =

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= lim

x∞[Ln(x + c) – Ln(x)]'

[x-1]'=

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= lim

x∞[Ln(x + c) – Ln(x)]'

[x-1]'=

lim x∞

1/(x + c) – 1/x-1/x2

=

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= lim

x∞[Ln(x + c) – Ln(x)]'

[x-1]'=

lim x∞

1/(x + c) – 1/x-1/x2

= lim x∞

x2 – x(x + c)-(x + 1)

=

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= lim

x∞[Ln(x + c) – Ln(x)]'

[x-1]'=

lim x∞

1/(x + c) – 1/x-1/x2

= lim x∞

x2 – x(x + c)-(x + 1)

= lim x∞

cxx + 1

=

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= lim

x∞[Ln(x + c) – Ln(x)]'

[x-1]'=

lim x∞

1/(x + c) – 1/x-1/x2

= lim x∞

x2 – x(x + c)-(x + 1)

= lim x∞

cxx + 1

= = c

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= lim

x∞[Ln(x + c) – Ln(x)]'

[x-1]'=

lim x∞

1/(x + c) – 1/x-1/x2

= lim x∞

x2 – x(x + c)-(x + 1)

= lim x∞

cxx + 1

= = c

Therefore lim (1 + c/x)x = ec.x∞

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

B. Find lim xx. (00 form)x0+

lim Ln(xx) = lim xLn(x) =x0+

So lim xx =x0+

lim eLn(x )x = 1

x0+

x0+

C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞

Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞

lim x∞

lim x∞

Ln(x + c) – Ln(x)

x-1= lim

x∞[Ln(x + c) – Ln(x)]'

[x-1]'=

lim x∞

1/(x + c) – 1/x-1/x2

= lim x∞

x2 – x(x + c)-(x + 1)

= lim x∞

cxx + 1

= = c

Therefore lim (1 + c/x)x = ec.x∞

lim Ln(x) x0+ x-1 =

L'Hopital

limx0+

x-1

-x-2 = 00

L'Hopital

–x

1

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞.

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x ( form, use L'Hopital's rule)

00

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x ( form, use L'Hopital's rule)

00

x0 [xsin(x)]'

[sin(x) – x]' = lim

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x ( form, use L'Hopital's rule)

00

x0 [xsin(x)]'

[sin(x) – x]' = lim

x0 sin(x)+ xcos(x)

cos(x) – 1= lim

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x ( form, use L'Hopital's rule)

00

x0 [xsin(x)]'

[sin(x) – x]' = lim

x0 sin(x)+ xcos(x)

cos(x) – 1= lim ( form, use L'Hopital's again)

00

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x ( form, use L'Hopital's rule)

00

x0 [xsin(x)]'

[sin(x) – x]' = lim

x0 sin(x)+ xcos(x)

cos(x) – 1= lim ( form, use L'Hopital's again)

00

x0 [sin(x)+ xcos(x)]'

[cos(x) – 1]'= lim

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x ( form, use L'Hopital's rule)

00

x0 [xsin(x)]'

[sin(x) – x]' = lim

x0 sin(x)+ xcos(x)

cos(x) – 1= lim ( form, use L'Hopital's again)

00

x0 [sin(x)+ xcos(x)]'

[cos(x) – 1]'= lim

x0 cos(x) + cos(x) – xsin(x)

-sin(x)= lim

D. Find lim ( )x0 x

L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)

The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.

1 sin(x)

1

lim( ) = x0 x

1 sin(x)

1 lim x0 xsin(x)

sin(x) – x ( form, use L'Hopital's rule)

00

x0 [xsin(x)]'

[sin(x) – x]' = lim

x0 sin(x)+ xcos(x)

cos(x) – 1= lim ( form, use L'Hopital's again)

00

x0 [sin(x)+ xcos(x)]'

[cos(x) – 1]'= lim

x0 cos(x) + cos(x) – xsin(x)

-sin(x)= lim = 0