Let Show that f(x) is discontinuous at x = 0.

__Ideas required to solve the problem:__

1. Meaning of continuity of function – If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x , f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x–coordinate of the point at which continuity is to be checked

If:–

equation 1

where h is a very small positive no (can assume h = 0.00000000001 like this )

It means :–

Limiting value of the left neighbourhood of x = c also called left hand limit LHL must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL and both must be equal to the value of f(x) at x=c i.e. f(c).

Thus, it is the necessary condition for a function to be continuous

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

2. Idea of sandwich theorem – This theorem also known as squeeze theorem that you may have encountered in your class 11 in limits chapter suggests that

If I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have{displaystyle lim _{x→ a}g(x)=lim _{x→ a}h(x)=L.}{displaystyle lim _{x→ a}f(x)=L.}

g(x) ≤ f(x) ≤ h(x) and also

Then , We say that f(x) is squeezed between g(x) and h(x) or you can assume it like sandwich.

∴ equation 2

NOTE : denominator in the above limit should be exactly same as that of content in sine function

Eg :

Let’s solve :

To check whether function is continuous at x=0 we need to check whether LHL = RHL = f(c)

As continuity is to be checked at x = 0 therefore c=0.

As, …… Equation 3

Clearly, f(0) = 1 using eqn 3

LHL =

Using equation 3 –

[∵ cos(–θ) = (cosθ)]

Whenever you see trigonometric term in numerator and corresponding or similar content of term in numerator try to apply sandwich theorem in finding apply if you are unable to do so think of alternative.

Here we have cos term but to apply the theorem we need to have sin term in numerator.

Can we bring it ?

Yes, we know that :

(1 – cos θ) = 2/2)

We have,

LHL = 2

To find its limit we need to think of Sandwich theorem as the form looks similar but term in denominator is not exactly same as that in the content of sine function.

Ok no problem, let’s bring it there but if we multiply 1/2 in denominator we need to multiply 1/2 in numerator too so that both can be cancelled.

Let’s do it:

∴ LHL =

= =

RHL =

Using equation 3 –

It is not taking or any other indeterminate form.

So, RHL =

Thus,

LHL ≠ RHL

∴ LHL ≠ RHL ≠ f(0)

∴ condition for function to be continuous is not satisfied

∴ f(x) is discontinuous at x = 0

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