limit-continuity-7

PROBLEMS (MEDIUM TYPE)

1) Without expansion or using L-Hospital’s Rule prove that .

Solution:

Let the limit be P then

Dumb Question:

1)      How can we replace q by 3q in the limit without changing it?

Ans: Since q ® 0, 3q will also tend to zero and thus there is no effect on the limit and thus it remains same.

2) Prove that

If x is rational.

If x is irrational.

Solution:

We know that |Cosq| £ 1 for all q.

 If |Cos (n!p x) | = P <1,

limit =

As P^2m ® 0 When m® .

If |Cos (n!p x) | =1.

limit =

Now |Cos (n!p x) | =1

Þ n!px = kp, Where k is an integer.

\x = which is a rational number.

\ From (2) limit = 2 if x is rational.

Again |Cos (n!p x) |<1

Þ |Cos (n!p x) | ¹ 1 for all n Î N.

Þ n!px ¹ kp for all nÎN, k Î Z

Þ x ¹ for all n Î N, k Î Z.

Þ x is not rational i.e. x is irrational.

\From (1) limit = 1 if x is irrational.

3) Given except at . Define so that f(x) may be continuous at .

Solution:

f (x) will be continuous at ,

If 

Dumb Question:

1)      Why did we multiply by in both numerator and denominator?

Ans: The basic aim to do the multiplication was to cancel out Cos2x which was making the limit an indeterminate form.

4) Prove that (where [] denotes greatest integer function) is continuous in .

Solution:

Here or , Where x = tanx ³ 0.

Then for a ÎN we discuss continuity of f(x) as L.H.L at x=a,

Now R.H.L at x=a.

So, g(x) is continuous for all a ÎN, Now g(x) is clearly continuous in (a-1, a) for all

a ÎN.

Hence g(x) is continuous in [0,].

Now let f(x) = tanx which is continuous in .

So g {f(x)} is continuous in .

Hence is continuous in .