Unit – 1
L’ Hospital’s Rules
Q1) What do you understand by indeterminate forms?
A1)
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 
Q2) What is L’Hospital rules for 0/0 indeterminate form?
A2)
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 can not 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).
Q3) Evaluate 
.
A3)
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
Q4) Evaluate 
A4)
Let f(x) = 
, then
And
= 0
= 0
But if we use L’Hospital rule again, then we get-
Q5) Evaluate 
A5)
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-
Q6) State and prove Cauchy’s mean value theorem.
A6)
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 satistfied 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.
Q7) Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]
A7)
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.
Q8) Examine the function 
for extreme values.
A8)
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
Q9) Prove that 
Then
A9)
Let f(x) = cos x
Then by Maclaurin’s series,
… (1)
Since
From Equation (1)
Q10) If 
Then
A10)
Here f(x) = sin hx.
By Maclaurin’s expansion,
(1)
By equation (1) we get,
Q11) If 
then
A11)
Here f(x) = log (1 + x)
By Maclaurin’s series expansion,
… (1)
By equation (1)
Q12) Expand by, Maclaurin’s theorem 
A12)
Here f(x) = log (1 + sin x)
By Maclaurin’s Theorem,
… (1)
……..
equation (1) becomes,
Q13) Expand by Maclaurin’s theorem, log sec x
A13)
Let f(x) = log sec x
By Maclaurin’s Expansion’s,
(1)
By equation (1)
Q14) Expand 
upto x6
A14)
Here 
Now we know that
… (1)
… (2)
Adding (1) and (2) we get
Q15) Expand 
by using Maclaurin’s series.
A15)
Let
Put these values in Maclaurin’s series-
Or
Q16) Expand 
in power of (x – 3)
A16)
Let 
Here a = 3
Now by Taylor’s series expansion,
… (1)
equation (1) becomes.
Q17) Expand 
in powers of x using Taylor’s theorem,
A17)
Here
i.e. 
Here
h = 2
By Taylors series
… (1)
By equation (1)
Q18) Find the value of 5
, correct to five decimal places by using the power series for 
.
A18)
As we know that the exponential series is-
Here we get-
Now
Q19) Expand 
as far as the term in 
.
A19)
We know that the power series for 
is-
Here we have to find- 
So that-
On solving, we get-
