Unit – 1A

Calculus

Q1) Consider the sequence . You might intuitively understand that this sequence converges to zero as n goes to infinity.  But we need to show this with a careful/rigorous mathematical argument.

A1) Formal argument (this is what you would want to write and turn in) Let Then when n> we have that

So, we have that < when n Therefore 0 as n

Q2) Show that converges to 1.

A2) We want First notice that no n gives a zero in the denomination. Now simplify = so that we need to show This is tricky because of the subtraction in the denominator. As long as n we have though. Now we just need to solve for n as a function of So as long as we choose a cut-off value the argument will work… Almost

Remember that this calculation originally needed n to keep the denominator positive. So we need satisfy  as well, and simultaneously satisfy  so that for any n we have n And n> We solve this by choosing some

Let and choose some

Then

Thus converges to 1.

Q3) p-series test:

(a) p   = .........

A3) Converges, if p>1

Diverges, if p<1

(b)  =

Here p = 3  so p>1 ,thus the given series converges.

=

Here p= ie., p<1, thus the given series diverges

Q4)

A4) +

= -3. +5.

=-3. +5.

Therefore., here p=2 ,3 ie., p>1

Hence the given series converges.

Q5) Find the limit of the following points

A5)

We apply the value of each x by finding the respective value through applying limits

f(x)=

f(x)=

Finding limit at x= 1.9

f(x)=

f(1.9)=

=

= 0.3448

Limit at x=1.99

f(1.99)=

=

= 0.33444

Limit at x=1.999

f(1.999)=

=

= 0.33344

Limit at x=2.001

f(2.001)=

=

= 0.33322

Limit at x=2.01

f(2.01)=

=

= 0.3225

From the above table we have to estimate the limit when x tends to 2

Here answer is 0.333....

Q6) Determine whether the following is dis-continuous at x=-1,0,

A6)

Given 

Verifying for continuity at x=-1

Therefore here = f(-1)

f(x)=f(-1)

Hence the function is continuous at x=-1

Verifying at x=0

Therefore here = f(0)

f(x)=f(0)

Hence the function is continuous at x=0

Q7)

A7) =

= (3) (

=

Q8) Consider the function f(x)= Discuss its continuity and differentiability at x=3/2

A8) For checking the continuity, we need to check the left hand and right-hand limits and the value of the function at a point x=a

L.H.L = R.H.L = F(a) =0

Thus, the function is continuous at about the point x=3/2

Now to check differentiability at the given point, we know

=

=

=

=

Thus, f is not differentiable at x=

Q9) Test the differentiability of the function

A9) We know that this function is continuous at x=2

But

Since the one-sided derivatives are not equal, does not exist. That is, f is not differentiable at x=2. At all other points, the function differentiable.

If is any other point then

Thus

The fact that does not exist is reflected geometrically in the fact that the curve does not have a tangent line at (2,0). Note that the curve has a sharp edge at (2,0).

Q10) Verify Rolle’s theorem for the function f(x) = x2 for

A10) Here f(x) = x2;

i)      Since f(x) is algebraic polynomial which is continuous in [-1, 1]

ii)    Consider f(x) = x2

Diff. w.r.t. x we get

f'(x) = 2x

Clearly f’(x) exists in (-1, 1) and does not becomes infinite.

iii)  Clearly

f(-1) = (-1)2 = 1

f(1) = (1)2 = 1

f(-1) = f(1).

Hence by Rolle’s theorem, there exist such that

f’(c) = 0

i.e. 2c = 0

c = 0

Thus, such that

f'(c) = 0

Hence Rolle’s Theorem is verified.

Q11) Verify Rolle’s Theorem for the function f(x) = ex (sin x – cos x) in

A11) Here f(x) = ex (sin x – cos x);

i)       Ex is an exponential function continuous for every also sin x and cos x are Trigonometric functions Hence (sin x – cos x) is continuous in and Hence ex(sin x – cos x) is continuous in .

ii)    Consider

f(x) = ex (sin x – cos x)

diff. w.r.t. x we get

f’(x) = ex(cos x + sin x) + ex(sin x + cos x)

= ex [2sin x]

Clearly f’(x) is exist for each & f’(x) is not infinite.

Hence f(x) is differentiable in .

iii)  Consider

Also,

Thus

Hence all the conditions of Rolle’s theorem are satisfied, so there exist such, that

i.e.

i.e. sin c = 0

But

Hence Rolle’s theorem is verified

Q12) Verify the Lagrange’s mean value theorem for

A12) Here

i)       Clearly f(x) = log x is logarithmic function. Hence it is continuous in [1, e]

ii)    Consider f(x) = log x.

Diff. w.r.t. x we get,

Clearly f’(x) exists for each value of & is finite.

Hence all conditions of LMVT are satisfied Hence at least

Such that

i.e.

i.e.

i.e.

i.e.

since e = 2.7183

Clearly c = 1.7183

Hence LMVT is verified.

Q13) Verify mean value theorem for f(x) = tan-1x in [0, 1]

A13) Here ;

i)       Clearly is an inverse trigonometric function and hence it is continuous in [0, 1]

ii)    Consider

diff. w.r.t. x we get,

Clearly f’(x) is continuous and differentiable in (0, 1) & is finite

Hence all conditions of LMVT are satisfied, Thus there exist

Such that

i.e.

i.e.

i.e.

i.e.

Clearly

Hence LMVT is verified.

Q14) Find the taylor series for the following:

= <1

A14) (x/10) <1 and (x/10) > -1

Therefore, radius of convergence is (-10,10)

  ROC =10

Q15) f(n)5 =

Here the ROC is 4

Compute the Taylor series centred at zero for f(x)= sinx

A15)

f(x)=sinx               f (0) = 0

f’(x)=cosx             f’ (0) =1

f’’(x)=-sinx           f’’ (0) =0

f’’’(x)= -cosx       f’’’ (0) =-1

f(4)(x) = cosx        f(4) (0)= 1

Applying Taylor series, we get

T(x) =   =   = x-

Thus, turns out to converge x to sinx.

Maclaurian series)

Q16)   (x)n

A16) f(x)=

= f(0)+f’(0)x+ x2 + x3 +......

= 1+x+x2 +x3 + .....

=

Q17) Find the Maclaurian series for f(x)= ex

A17) To get Maclaurian series, we look at the Taylor series polynomials for f near 0 and let them keep going.

Considering for example

By Maclaurian series we get,

+

Q18) Evaluate 

A18)

Differentiate the above form, we get

Now substitute the limit,

Therefore,

Q19) Evaluate

A19) Given,

Now substitute the limit

Therefore,

Q20) Find out the maxima and minima of the function

A20) Given     …(i)

Partially differentiating (i) with respect to x we get

    …. (ii)

Partially differentiating (i) with respect to y we get

    …. (iii)

Now, form the equations

Using (ii) and (iii) we get

          using above two equations

Squaring both sides, we get

Or   

This show that

Also, we get

Thus, we get the pair of value as

Now, we calculate

Putting above values in

At point (0,0) we get

So, the point (0,0) is a saddle point.

At point we get

So, the point is the minimum point where

In case 

So, the point is the maximum point where

Q21) Find the maximum and minimum point of the function

A21) Partially differentiating given equation with respect to and x and y then equate them to zero                 

On solving above, we get 

Also

Thus, we get the pair of values (0,0), (,0) and (0,

Now, we calculate

At the point (0,0)

So, function has saddle point at (0,0).

At the point (

So, the function has maxima at this point (.

At the point (0,

So, the function has minima at this point (0,.

At the point (

So, the function has an saddle point at (

Q22) Find the maximum and minimum value of

A22) Let

Partially differentiating given function with respect to x and y and equate it to zero

  ...(i)

    ...(ii)

On solving (i) and (ii) we get

Thus, pair of values are

Now, we calculate

At the point (0,0)

So further investigation is required

On the x axis y = 0, f(x,0) =0

On the line y=x,

At the point

So that the given function has maximum value at

Therefore, maximum value of given function

At the point

So that the given function has minimum value at

Therefore, minimum value of the given function

Q23)

A23) From the given equation

Q24)

A24)

= 49/4.