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 P2m ® 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 .