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.

1. 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-