Worked examples
The question
Question: find the limit as x → 4 of (√x − 2)/(x − 4).
lim x → 4 √x − 2x − 4 =
1 Put the number in
Put 4 in place of x.
The top is 0, and the bottom is 0.
√4 − 24 − 4 = 00
The square root of 4 is 2.
2 minus 2 is 0.
√4 = 2, so 2 − 2 = 0
0/0 has no value.
It tells us to work more.
00: no value yet. Work more!
2 Multiply by the conjugate
The conjugate of √x − 2 is √x + 2: the same two terms, with the sign swapped.
conjugate of √x − 2 is √x + 2
Multiply the top and the bottom by the conjugate.
lim x → 4 √x − 2x − 4 = lim x → 4 √x − 2x − 4 × √x + 2√x + 2
Something over itself is 1.
So the value does not change.
√x + 2√x + 2 = 1
Careful!
Both the top and the bottom!
One alone would change the value.
Careful: not only the top × (√x + 2) ✗
3 Multiply out
Remember: a minus b, times a plus b, is a squared minus b squared.
The middle terms cancel.
(a − b)(a + b) = a2 − b2
On top: (√x − 2) times (√x + 2).
That is the root squared, minus 2 squared.
The bottom we leave as it is.
= lim x → 4 (√x)2 − 22(x − 4)(√x + 2)
A square root squared is just what is inside.
So it is x − 4.
(√x)2 − 22 = x − 4
Now (x − 4) is on the top and on the bottom.
= lim x → 4 x − 4(x − 4)(√x + 2)
4 Cancel
x gets close to 4, but it is never 4.
So x − 4 is not 0.
x ≠ 4, so x − 4 ≠ 0
Something over itself is 1.
So we cross out (x − 4) on the top and the bottom.
Write what is left: 1 over (√x + 2).
= lim x → 4 1√x + 2
5 Put the number in again
Now the 0 is gone.
Put 4 in again.
= 1√4 + 2
The square root of 4 is 2.
= 12 + 2
So the limit is 1/4.
= 14
6 Check
Let's check with numbers close to 4.
A calculator gives 0.2502 at x = 3.99.
x = 3.99: f(x) ≈ 0.2502
And 0.2498 at x = 4.01.
Both are close to 1/4, which is 0.25.
Right!
x = 4.01: f(x) ≈ 0.2498
Let's look back.
First: put the number in.
Then: multiply by the conjugate.
Then: multiply out.
Then: cancel.
Then: put the number in again.
Last: check.
▶ Watch it solved
The question
Question: find the limit as x → 4 of (√(x + 5) − 3)/(x − 4).
lim x → 4 √x + 5 − 3x − 4 =
1 Put the number in
Put 4 in place of x.
The top is 0, and the bottom is 0.
√4 + 5 − 34 − 4 = 00
The square root of 9 is 3.
3 minus 3 is 0.
√9 = 3, so 3 − 3 = 0
0/0 has no value.
It tells us to work more.
00: no value yet. Work more!
2 Multiply by the conjugate
The conjugate of √(x + 5) − 3 is √(x + 5) + 3: the same two terms, with the sign swapped.
conjugate of √x + 5 − 3 is √x + 5 + 3
Multiply the top and the bottom by the conjugate.
lim x → 4 √x + 5 − 3x − 4 = lim x → 4 √x + 5 − 3x − 4 × √x + 5 + 3√x + 5 + 3
Something over itself is 1.
So the value does not change.
√x + 5 + 3√x + 5 + 3 = 1
Careful!
Both the top and the bottom!
One alone would change the value.
Careful: not only the top × (√(x + 5) + 3) ✗
3 Multiply out
Remember: a minus b, times a plus b, is a squared minus b squared.
The middle terms cancel.
(a − b)(a + b) = a2 − b2
On top: (√(x + 5) − 3) times (√(x + 5) + 3).
That is the root squared, minus 3 squared.
The bottom we leave as it is.
= lim x → 4 (√x + 5)2 − 32(x − 4)(√x + 5 + 3)
A square root squared is just what is inside.
So it is x − 4.
(√x + 5)2 − 32 = x − 4
Now (x − 4) is on the top and on the bottom.
= lim x → 4 x − 4(x − 4)(√x + 5 + 3)
4 Cancel
x gets close to 4, but it is never 4.
So x − 4 is not 0.
x ≠ 4, so x − 4 ≠ 0
Something over itself is 1.
So we cross out (x − 4) on the top and the bottom.
Write what is left: 1 over (√(x + 5) + 3).
= lim x → 4 1√x + 5 + 3
5 Put the number in again
Now the 0 is gone.
Put 4 in again.
= 1√4 + 5 + 3
The square root of 9 is 3.
= 13 + 3
So the limit is 1/6.
= 16
6 Check
Let's check with numbers close to 4.
A calculator gives 0.1667 at x = 3.99.
x = 3.99: f(x) ≈ 0.1667
And 0.1666 at x = 4.01.
Both are close to 1/6, which is 0.1667.
Right!
x = 4.01: f(x) ≈ 0.1666
Let's look back.
First: put the number in.
Then: multiply by the conjugate.
Then: multiply out.
Then: cancel.
Then: put the number in again.
Last: check.
▶ Watch it solved
The question
Question: find the limit as x → 8 of (√(x + 1) − 3)/(16 − 2x).
lim x → 8 √x + 1 − 316 − 2x =
1 Put the number in
Put 8 in place of x.
The top is 0, and the bottom is 0.
√8 + 1 − 316 − 2(8) = 00
The square root of 9 is 3.
3 minus 3 is 0.
√9 = 3, so 3 − 3 = 0
0/0 has no value.
It tells us to work more.
00: no value yet. Work more!
2 Multiply by the conjugate
The conjugate of √(x + 1) − 3 is √(x + 1) + 3: the same two terms, with the sign swapped.
conjugate of √x + 1 − 3 is √x + 1 + 3
Multiply the top and the bottom by the conjugate.
lim x → 8 √x + 1 − 316 − 2x = lim x → 8 √x + 1 − 316 − 2x × √x + 1 + 3√x + 1 + 3
Something over itself is 1.
So the value does not change.
√x + 1 + 3√x + 1 + 3 = 1
Careful!
Both the top and the bottom!
One alone would change the value.
Careful: not only the top × (√(x + 1) + 3) ✗
3 Multiply out
Remember: a minus b, times a plus b, is a squared minus b squared.
The middle terms cancel.
(a − b)(a + b) = a2 − b2
On top: (√(x + 1) − 3) times (√(x + 1) + 3).
That is the root squared, minus 3 squared.
The bottom we leave as it is.
= lim x → 8 (√x + 1)2 − 32(16 − 2x)(√x + 1 + 3)
A square root squared is just what is inside.
So it is x − 8.
(√x + 1)2 − 32 = x − 8
Take out the −2: 16 − 2x is −2 times (x − 8).
16 − 2x = −2(x − 8)
Now (x − 8) is on the top and on the bottom.
= lim x → 8 x − 8−2(x − 8)(√x + 1 + 3)
4 Cancel
x gets close to 8, but it is never 8.
So x − 8 is not 0.
x ≠ 8, so x − 8 ≠ 0
Something over itself is 1.
So we cross out (x − 8) on the top and the bottom.
Write what is left: −1/(2(√(x + 1) + 3)).
= lim x → 8 −12(√x + 1 + 3)
5 Put the number in again
Now the 0 is gone.
Put 8 in again.
= −12(√8 + 1 + 3)
The square root of 9 is 3.
= −12(3 + 3)
So the limit is −1/12.
= −112
6 Check
Let's check with numbers close to 8.
A calculator gives −0.0834 at x = 7.99.
x = 7.99: f(x) ≈ −0.0834
And −0.0833 at x = 8.01.
Both are close to −1/12, which is −0.0833.
Right!
x = 8.01: f(x) ≈ −0.0833
Let's look back.
First: put the number in.
Then: multiply by the conjugate.
Then: multiply out.
Then: cancel.
Then: put the number in again.
Last: check.
▶ Watch it solved
