MA 210 LECTURE (9)

Expansion by use of known series:

When the expansion of a function is required only up to first few terms, it is often
convenient to employ the following series:
𝜃3 𝜃5 𝜃7
1) sin 𝜃 = 𝜃 − + 5! − +⋯
3! 7!

𝜃3 𝜃5 𝜃7
2) sinh 𝜃 = 𝜃 + + 5! + +⋯
3! 7!

𝜃2 𝜃4 𝜃6
3) cos 𝜃 = 1 − 2! + − +⋯
4! 6!

𝜃2 𝜃4 𝜃6
4) cos 𝜃 = 1 + 2! + + +⋯
4! 6!

𝜃3 2𝜃5
5) tan 𝜃 = 𝜃 + + +⋯
3 15

𝑥3 𝑥5
6) tan−1 𝑥 = 𝑥 − + −⋯
3 5

𝑥2 𝑥3 𝑥4
7) 𝑒 𝑥 = 1 + 𝑥 + + + +⋯
2! 3! 4!

𝑥2 𝑥3 𝑥4
8) ln(𝑥 + 1) = 𝑥 − + − +⋯
2 3 4

𝑥2 𝑥3 𝑥4
9) ln(1 − 𝑥) = − [𝑥 + + + +⋯]
2 3 4

𝑛(𝑛−1) 𝑛(𝑛−1)(𝑛−2)
10) (1 + 𝑥)𝑛 = 1 + 𝑛𝑥 + 𝑥2 + 𝑥3 + ⋯
2! 3!

1
Example
Expand 𝑒 sin 𝑥 by Maclaurin’s series or otherwise up to the term containing 𝑥 4 .
Solution
We have;
(sin 𝑥)2 (sin 𝑥)3 (sin 𝑥)4
𝑒 sin 𝑥 = 1 + sin 𝑥 + + + +⋯
2! 3! 4!
2 3
𝑥3 1 𝑥3 1 𝑥3 1
= 1 + (𝑥 − + ⋯ ) + 2! (𝑥 − + ⋯ ) + 3! (𝑥 − + ⋯ ) + 4! (𝑥 − ⋯ )4 + ⋯
3! 3! 3!

𝑥3 1 𝑥4 1 1
= 1 + (𝑥 − + ⋯ ) + 2 (𝑥 2 − + ⋯ ) + 6 (𝑥 3 − ⋯ ) + 24 (𝑥 4 + ⋯ ) + ⋯
6 3
𝑥2 𝑥4
= 1+𝑥+ − +⋯
2! 8

Exercise
1) Expand ln(1 + sin2 𝑥) in powers of 𝑥 as far as the term in 𝑥 6 .

2) Prove that;

𝑥4 𝑥6
i) 𝑒 xsin 𝑥 = 1 + 𝑥 2 + + 120 + ⋯
3

𝑥2 𝑥3 𝑥4
ii) ln(1 + sin 𝑥) = 𝑥 − + − 12 + ⋯
2 6

𝑥 𝑥2 𝑥4
iii) ln(1 + 𝑒 𝑥 ) = ln 2 + 2 + − 192 + ⋯
8

Solution:
log( 1 + sin2 𝑥) = ln(1 + sin2 𝑥)
2
𝑥3 𝑥5
We have; sin2 𝑥 = (𝑥 − + −⋯)
3! 5!

𝑥4 𝑥6 𝑥4 𝑥6 𝑥6
= 𝑥2 − + 120 − + 36 + 120
6 6

𝑥4 2𝑥 6
= 𝑥2 − + + ⋯ = 𝑡, say.
3 45

𝑡2 𝑡3 𝑡4
Now ln(1 + sin2 𝑥) = ln(1 + 𝑡) = 𝑡 − + − +⋯
2 3 4

2
 Substituting the value of 𝑡 we get;
2
𝑥4 2𝑥 6 1 𝑥4 1
log( 1 + sin2 𝑥) = 𝑥 2 − + + ⋯ 2 (𝑥 2 − + ⋯ ) − 3 (𝑥 2 − ⋯ )3
3 45 3

𝑥4 2𝑥 6 1 2𝑥 6 1
= 𝑥2 − + − 2 (𝑥 4 − + ⋯ ) + 3 (𝑥 6 + ⋯ ) + ⋯
3 45 3

5 32
= 𝑥 2 − 6 𝑥 4 + 45 𝑥 6 + ⋯

Indeterminate forms
In general,

𝑓(𝑥) lim [𝑓(𝑥)]

𝑥→𝑎
lim [ ]=
𝑥→𝑎 𝑔(𝑥) lim [𝑔(𝑥)]
𝑥→𝑎

But when lim [𝑓(𝑥)] and lim [𝑔(𝑥)] are both zero, then the quotient reduces to the
𝑥→𝑎 𝑥→𝑎
0
indeterminate form 0 . This does not (mean) imply that
𝑓(𝑥)
lim [ ] is meaningless or it does not exist. In fact, in many cases, it
𝑥→𝑎 𝑔(𝑥)
has a finite value.
0
1) Form 0 : if 𝑓(𝑎) = 𝑔(𝑎) = 0, then

𝑓(𝑥) 𝑓 ′ (𝑥)
lim [𝑔(𝑥)] = lim [𝑔′ (𝑥)] (1)
𝑥→𝑎 𝑥→𝑎

This is known as 𝓛′ Hospital’s rule.

In general, if
𝑓(𝑎) = 𝑓 ′ (𝑎) = 𝑓 ′′ (𝑎) = ⋯ = 𝑓 𝑛−1 (𝑎) = 0, but 𝑓 𝑛 (𝑎) ≠ 0
𝑔(𝑎) = 𝑔′ (𝑎) = 𝑔′′ (𝑎) = ⋯ = 𝑔𝑛−1 (𝑎) = 0, but 𝑔𝑛 (𝑎) ≠ 0
Then from (1),
𝑓(𝑥) 𝑓𝑛 (𝑎) 𝑓𝑛 (𝑥)
lim [𝑔(𝑥)] = [𝑔𝑛(𝑎)] = lim [𝑔𝑛(𝑥)]
𝑥→𝑎 𝑥→𝑎

Note:

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 𝑓(𝑥) 0
To evaluate lim [𝑔(𝑥)] in 0 form, we differentiate the numerator and denominator
𝑥→𝑎
separately as many times as would be necessary to arrive at a determinate form.
Examples
Evaluate the following
𝑥𝑒 𝑥 −ln(1+𝑥)
i) lim
𝑥→0 𝑥2

𝑥 𝑥 −𝑥
ii) lim
𝑥→1 𝑥−1−ln 𝑥

Solutions
𝑥𝑒 𝑥 −ln(1+𝑥) 0
i) lim (form 0)
𝑥→0 𝑥2

1
(𝑥𝑒 𝑥 +𝑒 𝑥 ∙1)− 0
1+𝑥
= (form 0)
2𝑥

1
𝑥𝑒 𝑥 + 𝑒 𝑥 + 𝑒 𝑥 + 0+1+1+1 3
(1 + 𝑥)2
= = =
2 2 2

𝑥 𝑥 −𝑥 0
ii) lim (form 0)
𝑥→1 𝑥−1−ln 𝑥

Let 𝑦 = 𝑥 𝑥
ln 𝑦 = 𝑥 ln 𝑥
1 𝑑𝑦 1
∴ = 𝑥 ∙ 𝑥 + 1 ∙ ln 𝑥
𝑦 𝑑𝑥

𝑑
or (𝑥 𝑥 ) = 𝑥 𝑥 (1 + ln 𝑥) (𝑖)
𝑑𝑥

𝑑
(𝑥 𝑥 )−1
𝑑𝑥
= lim 1
𝑥→1 1−0−
𝑥

𝑥 𝑥 (1+ln 𝑥)−1
= lim 1
𝑥→1 1−
𝑥

𝑑 1
(𝑥 𝑥 )∙(1+ln 𝑥)+𝑥 𝑥 ( )−0
𝑑𝑥 𝑥
= lim 1
𝑥→1
𝑥2

1
𝑥 𝑥 (1+ln 𝑥)2 +𝑥 𝑥 ( ) 1(1+0)2 +1.1
𝑥
= lim = = 2.
𝑥→1 𝑥 −2 1

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 𝑥(𝑎+𝑏 cos 𝑥)−𝑐 sin 𝑥)
lim =1 Exam
𝑥→0 𝑥5

𝑥(𝑎+𝑏 cos 𝑥)−𝑐 sin 𝑥) 0

lim (form 0)
𝑥→0 𝑥5

𝑎+𝑏 cos 𝑥−𝑏𝑥 sin 𝑥−𝑐 cos 𝑥

lim (i)
𝑥→0 5𝑥 4

As the denominator is zero for 𝑥 = 0 (i) will tend

to a finite limit if and only if the numerator also becomes 0, for 𝑥 = 0. This
requires
0
𝑎 + 𝑏 − 𝑐 = 0 , (i) assumes the form 0
−𝑏 sin 𝑥−𝑏(sin 𝑥+𝑥 cos 𝑥)+𝑐 sin 𝑥
(i) = Lt
𝑥→0 20𝑥 3

(𝑐−2𝑏) sin 𝑥−𝑏𝑥 cos 𝑥 0

= Lt form 0
𝑥→0 20𝑥 3

(𝑐−2𝑏) cos 𝑥−𝑏 (cos 𝑥−𝑥 sin 𝑥)

= Lt (iii)
𝑥→0 60𝑥 2

𝑐−2𝑏−𝑏 𝑐−3𝑏
= = =1
0 0

∴ 𝑐 − 3𝑏 = 0 , i.e. 𝑐 = 3𝑏
𝑏 cos 𝑥−𝑏 cos 𝑥+𝑏𝑥 sin 𝑥 𝑏 sin 𝑥 𝑏 sin 𝑥
Now (iii) = Lt = Lt = 60 Lt
𝑥→0 60𝑥 2 𝑥→0 60𝑥 𝑥→0 𝑥

N.B:
∞
ℒ ′ Hospital’s rule can also be applied to the indeterminate form by
∞
differentiating the numerator and denominator separately as many times as would be
necessary.

5
Sometimes it is convenient to use expansion of known functions and standard limits for
evaluating the indeterminate forms. For this purpose, remember some well-known
series, then look-up and take note of the following standard limits:
sin 𝑥
(i) lim =1
𝑥→0 𝑥

1
(ii) lim(1 + 𝑥)𝑥 = 𝑒
𝑥→0

1 𝑥
(iii) lim (1 + 𝑥) = 𝑒
𝑛→∞

Exercise
1) Evaluate the following

ln(1−𝑥 2 )
(i) lim
𝑥→0 ln cos 𝑥

5 sin 𝑥−7 sin 2𝑥+3 sin 3𝑥

(ii) lim
𝑥→0 tan 𝑥−𝑥

𝑎𝑥 −1
(iii) lim
𝑥→0 𝑥

2𝑥 −1
(iv) lim 1
𝑥→0 (1+𝑥)2 −1

𝑥𝑛
(v) lim
𝑥→∞ 𝑒 𝑛

ln 𝑥
(vi) lim
𝑥→0 𝑐𝑜𝑠𝑒𝑐 𝑥

2) Find the values of 𝑎 and 𝑏 such that

𝑥(𝑎+𝑏 cos 𝑥)−𝑐 sin 𝑥

(i) lim =1
𝑥→0 𝑥5

𝑎 sinh 𝑥+𝑏 sin 𝑥 5

(ii) lim =3
𝑥→0 𝑥3

3) Find 𝑎, 𝑏, 𝑐 so that

𝑎𝑒 𝑥 −𝑏 cos 𝑥+𝑐 𝑒 −𝑥
lim 𝑥 sin 𝑥
=2
𝑥→0

6
 Evaluate
4)
2 sin 𝑥−sin 2𝑥
(i) lim
𝑥→0 𝑥3

1
ln sec 𝑥− 𝑥 2
2
(ii) lim
𝑥→0 𝑥4

𝑒 𝑥 +2 sin 𝑥−𝑒 −𝑥 −4𝑥

(iii) lim
𝑥→0 𝑥5

𝑒 𝑥 +sin 𝑥−1
(iv) lim
𝑥→0 ln(1+𝑥)

Increasing and decreasing functions

In the function 𝑦 = 𝑓(𝑥) , if 𝑦 increases as 𝑥 increases (𝑎𝑠 𝑎𝑡 𝐴) , it is called an
increasing function of 𝑥. On the other hand, if 𝑦 decreases ( 𝑎𝑠 𝑎𝑡 𝑐), as 𝑥
increases, it is called a decreasing function of 𝑥.
Let the tangent at any point on the graph of the function make an angle of 𝜃 with
𝑑𝑦
the 𝑥 − 𝑎𝑥𝑖𝑠 (figure below) so that 𝑑𝑥 = tan 𝜃.

y
𝐵
𝑦 = 𝑓(𝑥)

A C 𝐷

𝜃 𝜃
x

𝑑𝑦
At any point such as A, where the function is increasing, < 𝜃 is acute i.e.,
𝑑𝑥
is positive. (skewed tangent to the right)
𝑑𝑦
At any point C, Where the function is decreasing, < 𝜃 is obtuse i.e. 𝑑𝑥 is
negative (skewed tangent to the left).
Hence the derivative of an increasing function is +ve, and the derivative of a
decreasing function is – ve.
If the derivative is zero (as at B or D), then 𝑦 is neither increasing nor
decreasing. In such cases, we say that the function is stationary.

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Concavity, convexity and point of inflection
(i) If a portion of the curve on both sides of a point, however small it may be, lies
above the tangent (as at D), then the curve is said to be concave downwards at
𝑑2 𝑦
D where 𝑑𝑥 2 is positive (minimum curve)
(ii) If a portion of the curve on both sides of a point, lie below the tangent (as at B),
𝑑2 𝑦
then the curve is said to be convex upwards at B where is negative
𝑑𝑥 2
(maximum)
(iii) If the two portions of the curve lie on different sides of the tangent (i.e., the
curve crosses the tangent (as at c), then the point c is said to be a point of
inflection of the curve.
𝑑2 𝑦 𝑑3 𝑦
At a point of inflection = 0 and 𝑑𝑥 3 ≠ 0
𝑑𝑥 2

Maxima and Minima

Consider the graph of the continuous function 𝑦 = 𝑓(𝑥) in the interval (𝑥1 , 𝑥2 );

Y 𝑃3
𝑃1

𝑃4
𝑥 = 𝑥2

𝑃2
𝑥 = 𝑥1

𝑥=𝑎

x
𝐿1 𝐿2

8
Clearly, the point 𝑝1 is the highest in its own immediate neighborhood. So also, is 𝑃3 .At
each of these points, 𝑃1 , 𝑃3 , the function is said to have a maximum value.
On the other hand, the point 𝑝2 is the lowest in its own immediate neighborhood. So
also, is 𝑝4 . At each of these points 𝑝2 , 𝑝4 the function is said to have a minimum value.
Definition
A function 𝑓(𝑥) is said to have a maximum at 𝑥 = 𝑎, if there exists a small number h,
however small, such that 𝑓(𝑎) > both 𝑓(𝑎 − ℎ) and 𝑓(𝑎 + ℎ).
A function 𝑓(𝑥) is said to have a minimum value at 𝑥 = 𝑎, if there exists a small number
h, however small, such that 𝑓(𝑎) < both 𝑓(𝑎 − ℎ) and 𝑓(𝑎 + ℎ)
If a function is maximum or minimum at 𝑥 = 𝑎, then
𝑑𝑦
=0
𝑑𝑥 𝑥=𝑎

𝑑𝑦
In the interval (𝑎 − ℎ, 𝑎) before 𝐿1 , 𝑑𝑥 ≥ 0; at 𝑥 = 𝑎
𝑑𝑦 𝑑𝑦
= 0 and in the interval (𝑎 + ℎ, 𝑎), 𝑑𝑥 ≤ 0. (𝑖. 𝑒. 𝑎𝑓𝑡𝑒𝑟 𝐿1 )
𝑑𝑥

If 𝑓 ′ (𝑎) = 0 and 𝑓 ′′ (𝑎) is -ve (i.e., 𝑓 ′ (𝑎) changes sign from +ve to -ve), then
𝑓(𝑥) is maximum at 𝑥 = 𝑎
If 𝑓 ′ (𝑎) = 0 and 𝑓 ′′ (𝑎) is +ve (i.e., 𝑓 ′ (𝑎) changes sign from -ve to +ve), then
𝑓(𝑥) is minimum at 𝑥 = 𝑎.
Note:
𝑓 ′′ (𝑥) may sometimes be zero at 𝑥 = 𝑎, hence making it difficult to determine the
nature of the function. If that’s the case, see if (check) 𝑓 ′ (𝑥) changes sign from +ve to
-ve as 𝑥 passes through 𝑎, then 𝑓(𝑥) is maximum at 𝑥 = 𝑎. If 𝑓 ′ (𝑥) changes from -ve
to +ve as 𝑥 passes through 𝑎, 𝑓(𝑥) is minimum at 𝑥 = 𝑎. If 𝑓 ′ (𝑥) does not change sign
while passing through 𝑥 = 𝑎, 𝑓(𝑥) is neither minimum nor maximum at 𝑥 = 𝑎.
Example 1
Find the maximum and minimum values of 3𝑥 4 − 2𝑥 3 − 6𝑥 2 + 6𝑥 + 1
In the interval (0,2).
Solutions
let 𝑓(𝑥) = 3𝑥 4 − 2𝑥 3 − 6𝑥 2 + 6𝑥 + 1
𝑓 ′ (𝑥) = 12𝑥 3 − 6𝑥 2 + 12𝑥 + 6 = 6(𝑥 2 − 1)(2𝑥 − 1)
1
𝑓 ′ (𝑥) = 0 when 𝑥 = ±1, 2 ∙

9
 1
∴ 𝑓(𝑥) can have a maximum or minimum in the interval (0,2) at 𝑥 = 2 𝑜𝑟 𝑥 = 1
1
Now 𝑓 ′′ (𝑥) = 36𝑥 2 − 12𝑥 − 12 = 12(3𝑥 2 − 𝑥 − 1) so that 𝑓 ′′ (2) = −9

And 𝑓 ′′ (1 = 12)
1
∴ 𝑓(𝑥) can have a maximum at 2 and a minimum at 𝑥 = 1.

Thus, the minimum value is;

1 1 4 1 3 1 2 1 7
𝑓( ) = 3( ) − 2( ) − 6( ) + 6( ) + 1 = 2
2 2 2 2 2 16
And the minimum value is;
𝑓(1) = 3(1)4 − 2(1)3 − 6(1)2 + 6(1) + 1

Example 2
𝜋
Show that sin 𝑥(1 + cos 𝑥) is a maximum when 𝑥 = 3

Solution
Let 𝑓(𝑥) = sin 𝑥(1 + cos 𝑥)
𝑓 ′ (𝑥) = cos 𝑥(1 + cos 𝑥) + sin 𝑥(− sin 𝑥)
= cos 𝑥(1 + cos 𝑥) − (1 − cos2 𝑥) = (1 + cos 𝑥)(2 cos 𝑥 − 1)
1 𝜋
𝑓 ′ (𝑥) = 0 when cos 𝑥 = 2 or −1 i.e., when 𝑥 = or 𝜋
3

Now 𝑓 ′′ (𝑥) = − sin 𝑥(2 cos 𝑥 − 1) + (1 + cos 𝑥)(−2 sin 𝑥) = − sin 𝑥(4 cos 𝑥 + 1)
𝜋 −3√2
So that 𝑓 ′′ ( 3 ) = and 𝑓 ′′ (𝜋) = 0
2
𝜋
Thus, 𝑓(𝑥) has a maximum at 𝑥 = 3

Since 𝑓 ′′ (𝜋) = 0 , let us check whether 𝑓 ′ (𝑥) changes sign or not.

When 𝑥 is slightly < 𝜋 , 𝑓 ′ (𝑥) is -ve.
When 𝑥 is slightly > 𝜋 , 𝑓 ′ (𝑥) is -ve (again). Therefore, 𝑓 ′ (𝑥) does not
change sign as 𝑥 passes through 𝜋. So that 𝑓(𝑥) is neither maximum nor minimum
at 𝑥 = 𝜋.

10
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