Unit - 1
L’ Hospital’s Rules
Let we have two functions f(x) and g(x) and-
Then-
Is an expression of the form 
.
In that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Now, Let we have two functions f(x) and g(x) and-
Then-
Is an expression of the form 
, in that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Some other indeterminate forms are 
L’ Hospital’s rule for 
form-
Working steps-
1. Check that the limits f(x)/g(x) is an indeterminate form of type 
.
(Note- we cannot apply L’ Hospital rule if it is not in indeterminate form)
2. Differentiate f and g separately.
3. Find the limits of the derivatives. If the limit is finite,
then it is equal to the limit of f(x)/g(x).
Example-1: Evaluate 
Sol. Here we notice that it is an indeterminate form of 
.
So that, we can apply L’ Hospital rule-
Example-2: Evaluate 
.
Sol. Let f(x) = 
and g(x) = 
.
Here we see that this is the indeterminate form of 0/0 at x = 0.
Now by using L’ Hospital rule, we get-
= 
= 
= 
= 1
Note- Suppose we get an indeterminate form even after finding first derivative, then in that case, we use the other form of L’ Hospital’s rule.
If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞, -∞), then
Example-3: Evaluate 
Sol.
Let f(x) = 
, then
And
= 0
= 0
But if we use L’ Hospital rule again, then we get-
Example-4: Evaluate 
Sol. We can see that this is an indeterminate form of type 0/0.
Apply L’ Hospital’s rule, we get
But this is again an indeterminate form, so that we will again apply L’ Hospital’s rule-
We get
= 
L’ Hospital’s rule for 
form-
Let f and g are two differentiable functions on an open interval containing x = a, except possibly at x = a and that
If 
has a finite limit, or if it is 
, then
Theorem- If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞, -∞), then
Example-5: Find 
, n>0.
Sol. Let f(x) = log x and g(x) = 
These two functions satisfied the theorem that we have discussed above-
So that,
Example-6: Evaluate 
Sol.
Apply L’ Hospital rule as we can see that this is the form of 
= 
Note- In some cases like above example, we cannot apply L’ Hospital’s rule.
Other types of indeterminate forms-
Example-7: Evaluate 
Sol. Here we find that-
So that this limit is the form of 0
.
Now,
Change 
to obtain the limit-
Now this is the form of 0/0,
Apply L’ Hospital’s rule-
Key takeaways-
If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞, -∞), then
Cauchy’s mean value theorem-
Suppose we have two functions f(x) and g(x) of x, such that,
1. Both functions are continuous in [a, b]
2. Both functions are differentiable in (a, b)
3. g’(x) ≠ 0 for any x ϵ (a, b)
These three exists atleast, x = c ϵ (a, b), at which
Proof:
Suppose, we define a functions,
h(x) = f(x) – A.g(x) …………………….(1)
So that h(a) = h(b) and A is a constant to be determined.
Now,
h(a) = f(a) – Ag(a)
h(b) = f(b) – Ag(b)
So that,
f(a) – Ag(a) = f(b) – A.g(b),
Which gives
A = 
…………………………….(2)
Now, h(x) is continuous in [a, b] as RHS of eq. (1) is continuous in [a, b] and h(x) is diff. In (a, b) as RHS of eq. (1) is diff. In (a, b)
Also,
h(a) = h(b)
Therefore all the conditions of Rolle’s theorem are satisfied then there exists a Value x = cϵ (a, b)
So that h’(c) = 0
Differentiate eq.(1) w.r.t. x , we get
h’(x) = f’(x) – A.g’(x)
At x = c
h’(c) = f’(c) – A.g’(c)
0 = f’(c) – A.g’(c)
A = 
So that, we get
where a<c<b
Hence the Cauchy’s mean value theorem is proved.
Example: Verify Cauchy’s mean value theorem for the function f(x) = sin x and g(x) = cosx in [ 0, π/2]
Sol. It is given that,
f(x) = sin x and g(x) = cos x
Now,
f’(x) = cos x and g’(x) = - sin x
We know that both the functions are continuous in [ 0, π/2] and differentiable in (0, π/2).
Also, g’(x) = - sin x ≠ 0 for all x ϵ (0, π/2)
By Cauchy’s mean value theorem, we get
for some c: 0< c <
That means
which gives,
Cot c = 1
c = 
Now we see that 
lies between 0 and 
Example: Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]
Sol. We are given, f(x) = x⁴ and g(x) = x
Derivative of these functions,
f’(x) = 4x³ and g’(x) = 2x
Put these values in Cauchy’s formula, we get
2c² = 
c² = 
c = 
Now put the values of a = 1 and b = 2, we get
c = 
=
= 
(approx)
Hence the Cauchy’s theorem is verified.
Key takeaways
Suppose we have two functions f(x) and g(x) of x, such that,
1. Both functions are continuous in [a, b]
2. Both functions are differentiable in (a, b)
3. g’(x) ≠ 0 for any x ϵ (a, b)
These three exists atleast, x = c ϵ (a, b), at which
If a function f is defined on [a, a + h] is such that
The (n-1)th derivative 
is continuous on [a a+h] and (ii) the nth derivative 
exists on ] a, a+h [ then for every for all at least one real number 
between 0 and 1, such that
Where p is the given positive integer.
We see that the condition (i) in the statement implies all the derivatives f’ f’’….. And are continuous on [a, a + h]
Consider the function 
defined on [a, a + h] as
Where A is a constant to be determined such that
The function satisfies all the conditions of Rolle’s theorem and hence for every for all at least one real number 
between 0 to 1, such that
But
We get
The term
Which occurs after n terms is called Taylor’s remainder.
Cauchy form of remainder
P=1, we get
Lagrange’s form of remainder
P=n, we get
Suppose c is the interior point of the domain [a, b] of a function f. Then the point c is said to be stationary if f’(c) = 0.
Maximum value- The function f is said to have a maximum value at c if f(c) is the greatest value of the function in a small neighbourhood c – 
, c +
[,
of c.
Thus
We have
Such that, f(x) – f(c) is negative for all values of x in c – 
, c +
[ other than c.
Minimum value- The function f is said to have a minimum value at c if f(c) is the least value of the function in a small neighbourhood c – 
, c +
[,
of c.
Thus
We have
Such that, f(x) – f(c) is positive for all values of x in ] c – 
, c +
[ other than c.
Extreme value- The function is said to have an extreme value at c if f(c) is either a maximum or a minimum value.
Thus at an extreme point c, f(x) – f(c) keeps the same sign for all values of x is a deleted neighborhood ] c – 
, c +
[ of c.
Method to find out the points of maximum and minimum values
Let c be an extreme point of a function f with domain [a, b]. If c is a point of maximum value, then for every for all a neighborhood ] c – 
, c +
[ of c such that
So that
f is increasing from f(c) in a small interval to the right of c.
We can conclude that
C is a point of maximum value if the function changes from an increasing to a decreasing function as x passes through c.
Therefore if f is derivable, the derivative changes sign from positive to negative as x passes through c.
It is clear that c is a point of minimum value if f changes from decreasing to an increasing function and in case f is derivable, the derivative f’ changes sign from negative to positive as x passes through c.
Theorem- if c is an interior point of the domain of a function and f’(c) = 0, then the function has a maximum or a minimum at c as f’’(c) is negative or positive.
F’’(c) implies that f and f’ exists and are continuous at c and continuity at c implies the existence of f and f’ in a neighbourhood ] c – 
, c +
[,
of c, the neghbourhood being itself contained in a domain D of f.
Suppose
This implies that f’(x) is an increasing function atc,
And
First one implies that f’(x) is positive and hence f is an increasing function in ] c, c + 
[ .
Similarly, f(x) > f(c) in 
These last results imply that
Has a minima at c.
Similarly
Note- Extreme values exists only if the first non-zero derivative is of even order.
Example: Examine the function 
for extreme values.
Sol.
Suppose
Then
Hence the function is derivable for all x 
and the derivative f’ vanishes for x = -1, 3 and 7/9 which we may test for extreme values.
- For x = -1
f’ is positive for a value of x less than -1 and negative for greater than -1.
Thus f’ changes sign from + to – as x passes through -1.
Hence -1 is a point of maximum
2. X = 3
F’ remains positive as x passes through 3.
So that x= 3 is
Key takeaways
- Minimum value- The function f is said to have a minimum value at c if f(c) is the least value of the function in a small neighbourhood ] c –
, c +
[,
of c. - Extreme value- the function is said to have an extreme value at c if f(c) is either a maximum or a minimum value
- If c is an interior point of the domain of a function and f’(c) = 0, then the function has a maximum or a minimum at c as f’’(c) is negative or positive.
- Extreme values exists only if the first non-zero derivative is of even order.
Maclaurin’s Series Expansions
Statement:-
Maclaurin’s series of f(x) at x = 0 is given by,
Expansion of some standard functions
i) f(x) = ex then
Proof:-
Here
By Maclaurin’s series we get,
i.e. 
Note that
- Replace x by –x we get
2. f(x) = sin x then
Proof:
Let (x) = sin x
Then by Maclaurin’s series,
… (1)
Since
By equation (i) we get,
3. 
Then
Proof:
Let f(x) = cos x
Then by Maclaurin’s series,
… (1)
Since
From Equation (1)
4. 
then
Proof:
Here f(x) = tan x
By Maclaurin’s expansion,
… (1)
Since
…..
By equation (1)
5. 
Then
Proof:-
Here f(x) = sin hx.
By Maclaurin’s expansion,
(1)
By equation (1) we get,
6. 
. Then
Proof:-
Here f(x) = cos hx
By Maclaurin’s expansion
(1)
By equation (1)
7. f(x) = tan hx
Proof:
Here f(x) = tan hx
By Maclaurin’s series expansion,
… (1)
By equation (1)
8. 
then
Proof:-
Here f(x) = log (1 + x)
By Maclaurin’s series expansion,
… (1)
By equation (1)
9. 
In above result we replace x by -x
Then
10. Expansion of tan h-1x
We know that
Thus
11. Expansion of (1 + x)m
Proof:-
Let f(x) = (1 + x)m
By Maclaurin’s series.
… (1)
By equation (1) we get,
Note that in above expansion if we replace m = -1 then we get,
Now replace x by -x in above we get,
Expand by, Maclaurin’s theorem
Solution:
Here f(x) = log (1 + sin x)
By Maclaurin’s Theorem,
… (1)
……..
equation (1) becomes,
Expand by Maclaurin’s theorem,
Log sec x
Solution:
Let f(x) = log sec x
By Maclaurin’s Expansion’s,
(1)
By equation (1)
Example: Prove that 
Solution:
Here f(x) = x cosec x
= 
Now we know that
Example: Expand 
upto x6
Solution:
Here 
Now we know that
… (1)
… (2)
Adding (1) and (2) we get
Example: Show that 
Solution:
Here 
Thus
Expand 
by using Maclaurin’s series. C
Sol.
Let
Put these values in Maclaurin’s series-
Or
Taylor’s Series Expansion:
a) The expansion of f(x+h) in ascending power of x is
b) The expansion of f(x+h) in ascending power of h is
c) The expansion of f(x) in ascending powers of (x-a) is,
Using the above series expansion we get series expansion of f(x+h) or f(x).
Expansion of functions using standard expansions
Example: Expand 
in power of (x – 3)
Solution:
Let 
Here a = 3
Now by Taylor’s series expansion,
… (1)
equation (1) becomes.
Example: Using Taylors series method expand
in powers of (x + 2)
Solution:
Here 
a = -2
By Taylors series,
… (1)
Since
, 
, …..
Thus equation (1) becomes
Example: Expand 
in ascending powers of x.
Solution:
Here
i.e. 
Here h = -2
By Taylors series,
… (1)
equation (1) becomes,
Thus
Example: Expand 
in powers of x using Taylor’s theorem,
Solution:
Here
i.e. 
Here
h = 2
By Taylors series
… (1)
By equation (1)
Key takeaways-
- Maclaurin’s Series Expansions
2. Taylor’s Series Expansion:-
Let’s first define the exponential functions-
A function which contains 
where e is the constant, called an exponential function.
Here ‘e’ is the exponent which has an approximate value 2.7183
The power series for 
The value of 
can be defined in terms of the following power series:
If we obtain an actual value of 
by adding all the terms of the series, then the series is said to be convergent.
The more terms we take in the series, the closer will be the value of 
to its actual value.
Example: Find the value of 5
, correct to five decimal places by using the power series for 
.
Sol. As we know that the exponential series is-
Here we get-
Now
Example: Expand 
as far as the term in 
.
Sol. We know that the power series for 
is-
Here we have to find- 
So that-
On solving, we get-
Trigonometric and log function-
As we know that,
And
By adding these equations-
And subtracting (2) from (1), we get-
Note- 
Example-1: Verify 
Sol.
As we know that
And
Hence,
And
since 
That means
The following function is a logarithmic function-
‘a’ is any value which is greater than 0, except 1
The natural logarithmic function is-
The value of
e = 2.71828182….
Properties of logarithms-
1.
2. 
3. 
4. 
then x = y
Example: Simply the logarithm 
Sol. We can write this as-
= 3
Example: Simplify 
Sol. We know that-
So that
Key takeaways
- A function which contains
where e is the constant, called an exponential function. 
References:
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. Advanced engineering mathematics, by HK Dass
3. Mathematical analysis by Dr. Anju panwar
4. Real analysis by Dr. Sachin Kaushal
5. Real analysis by Ajit kumar & S.Kumaresan
6. Principles of real analysis by S.C Malik
