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L'Hôpital's rule

Calculus 1 · 24 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Put the number in
  2. Use L'Hôpital's rule
  3. Use it again
  4. Put the number in again
  5. Check

The steps for Limit as x -> 0 of (1 - cos(x))/x^2. Every solution names its steps, so you always know what comes next.

Worked examples

Limit as x -> 0 of sin(x)/x Easy

The question

Question: find the limit as x → 0 of sin x over x.

lim x → 0 sin xx =

1 Put the number in

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: sin 0 = 0
bottom: 0

0/0 has no value.

It tells us nothing yet.

This is where L'Hôpital's rule helps.

form 00

2 Use L'Hôpital's rule

The rule: when we get 0/0, or infinity over infinity, we may take the derivative of the top, and the derivative of the bottom.

The new limit is the same.

lim x → a f(x)g(x) = lim x → a f′(x)g′(x)

Careful!

This is not the quotient rule.

The top gets its own derivative, and the bottom gets its own derivative.

Careful: not the quotient rule: top and bottom each alone

The derivative of the top is cos x.

The derivative of the bottom is 1.

The new limit is the same.

lim x → 0 sin xx = lim x → 0 cos x1

3 Put the number in again

Put 0 in again.

Now the bottom is not 0.

So the limit is 1.

= cos 01 = 11 = 1

4 Check

Let's check with a calculator in radians, with x very close to 0.

x = −0.01 gives about 1.

x = −0.01: sin xx ≈ 1

And x = 0.01 gives about 1.

Both are close to 1.

Right!

x = 0.01: sin xx ≈ 1

Let's look back.

First: put the number in.

Then: use L'Hôpital's rule.

Then: put the number in again.

Last: check.

Answer
= cos 01 = 11 = 1

▶ Watch it solved

Limit as x -> 0 of (e^(2x) - 1)/sin(x) Medium

The question

Question: find the limit as x → 0 of (e²ˣ − 1) over sin x.

lim x → 0 e2x − 1sin x =

1 Put the number in

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: e0 − 1 = 0
bottom: sin 0 = 0

0/0 has no value.

It tells us nothing yet.

This is where L'Hôpital's rule helps.

form 00

2 Use L'Hôpital's rule

The rule: when we get 0/0, or infinity over infinity, we may take the derivative of the top, and the derivative of the bottom.

The new limit is the same.

lim x → a f(x)g(x) = lim x → a f′(x)g′(x)

Careful!

This is not the quotient rule.

The top gets its own derivative, and the bottom gets its own derivative.

Careful: not the quotient rule: top and bottom each alone

The derivative of the top is 2e²ˣ.

The derivative of the bottom is cos x.

The new limit is the same.

lim x → 0 e2x − 1sin x = lim x → 0 2e2xcos x

3 Put the number in again

Put 0 in again.

Now the bottom is not 0.

So the limit is 2.

= 2e0cos 0 = 21 = 2

4 Check

Let's check with a calculator in radians, with x very close to 0.

x = −0.01 gives about 1.9802.

x = −0.01: e2x − 1sin x ≈ 1.9802

And x = 0.01 gives about 2.0202.

Both are close to 2.

Right!

x = 0.01: e2x − 1sin x ≈ 2.0202

Let's look back.

First: put the number in.

Then: use L'Hôpital's rule.

Then: put the number in again.

Last: check.

Answer
= 2e0cos 0 = 21 = 2

▶ Watch it solved

Limit as x -> 0 of (1 - cos(x))/x^2 Harder

The question

Question: find the limit as x → 0 of (1 − cos x)/x².

lim x → 0 1 − cos xx2 =

1 Put the number in

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: 1 − cos 0 = 0
bottom: 02 = 0

0/0 has no value.

It tells us nothing yet.

This is where L'Hôpital's rule helps.

form 00

2 Use L'Hôpital's rule

The rule: when we get 0/0, or infinity over infinity, we may take the derivative of the top, and the derivative of the bottom.

The new limit is the same.

lim x → a f(x)g(x) = lim x → a f′(x)g′(x)

Careful!

This is not the quotient rule.

The top gets its own derivative, and the bottom gets its own derivative.

Careful: not the quotient rule: top and bottom each alone

The derivative of the top is sin x.

The derivative of the bottom is 2x.

The new limit is the same.

lim x → 0 1 − cos xx2 = lim x → 0 sin x2x

3 Use it again

Put 0 in place of x.

The top is 0, and the bottom is 0.

top: sin 0 = 0
bottom: 2(0) = 0

0/0 again!

So we use the rule again.

form 00

The derivative of the top is cos x.

The derivative of the bottom is 2.

The new limit is the same.

= lim x → 0 cos x2

4 Put the number in again

Put 0 in again.

Now the bottom is not 0.

So the limit is 1/2.

= cos 02 = 12 = 12

5 Check

Let's check with a calculator in radians, with x very close to 0.

x = −0.01 gives about 0.5.

x = −0.01: 1 − cos xx2 ≈ 0.5

And x = 0.01 gives about 0.5.

Both are close to 1/2.

Right!

x = 0.01: 1 − cos xx2 ≈ 0.5

Let's look back.

First: put the number in.

Then: use L'Hôpital's rule.

Then: use it again.

Then: put the number in again.

Last: check.

Answer
= cos 02 = 12 = 12

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. limit as x -> 0 of sin(x)/x
    Answer
    = cos 01 = 11 = 1
  2. limit as x -> 0 of sin(3x)/x
    Answer
    = 3cos 01 = 31 = 3
  3. limit as x -> 0 of (e^x - 1)/x
    Answer
    = e01 = 11 = 1
  4. limit as x -> 0 of x/(e^x - 1)
    Answer
    = 1e0 = 11 = 1
  5. limit as x -> 1 of ln(x)/(x - 1)
    Answer
    = 111 = 11 = 1
  6. limit as x -> 0 of (e^(3x) - 1)/x
    Answer
    = 3e01 = 31 = 3
  7. limit as x -> 2 of (x^3 - 8)/(x - 2) using L'Hôpital's rule
    Answer
    = 3(2)21 = 121 = 12
  8. limit as x -> 3 of (x^2 - 9)/(x - 3) using L'Hôpital's rule
    Answer
    = 2(3)1 = 61 = 6

Medium

  1. limit as x -> 0 of (e^(2x) - 1)/sin(x)
    Answer
    = 2e0cos 0 = 21 = 2
  2. limit as x -> 0 of sin(5x)/sin(2x)
    Answer
    = 5cos 02cos 0 = 52 = 52
  3. limit as x -> 1 of (x^2 - 1)/ln(x)
    Answer
    = 2(1)11 = 21 = 2
  4. limit as x -> 1 of (x^4 - 1)/(x^2 - 1) using L'Hôpital's rule
    Answer
    = 4(1)32(1) = 42 = 2
  5. lim x->0 (e^x - cos(x))/x
    Answer
    = e0 + sin 01 = 11 = 1
  6. limit as x -> infinity of ln(x)/x
    Answer
    = 0
  7. limit as x -> infinity of x/e^x
    Answer
    = 0
  8. limit as x -> infinity of (3x^2 + 1)/(2x^2 - x) using L'Hôpital's rule
    Answer
    = 32
  9. limit as x -> 1 of (x^2 + 5)/(3x + 4) using L'Hôpital's rule
    Answer
    lim x → 1 x2 + 53x + 4 = 67 = 67
  10. limit as x -> 0 of (x^2 + 2)/(x + 1) using L'Hôpital's rule
    Answer
    lim x → 0 x2 + 2x + 1 = 21 = 2

Harder

  1. limit as x -> 0 of (1 - cos(x))/x^2
    Answer
    = cos 02 = 12 = 12
  2. limit as x -> 0 of (e^x - 1 - x)/x^2
    Answer
    = e02 = 12 = 12
  3. limit as x -> 0 of (1 - cos(2x))/x^2
    Answer
    = 4cos 02 = 42 = 2
  4. limit as x -> infinity of x^2/e^x
    Answer
    = 0
  5. limit as x -> infinity of e^x/x^2
    Answer
    = ∞
  6. limit as x -> infinity of ln(x)/x^2
    Answer
    = 0