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Question: By the definition of continuity, how do you show that \[x\sin \left( \dfrac{1}{x} \right)\] is conti...

By the definition of continuity, how do you show that xsin(1x)x\sin \left( \dfrac{1}{x} \right) is continuous at x=0x=0?

Explanation

Solution

From the question given, we have been asked to show xsin(1x)x\sin \left( \dfrac{1}{x} \right) is continuous at x=0x=0 by using the definition of continuity. By using the definition of continuity, to show that xsin(1x)x\sin \left( \dfrac{1}{x} \right) is continuous at x=0x=0, first of all, we have to know the definition of continuity.

Complete step by step solution:
For answering this question we need to show that xsin(1x)x\sin \left( \dfrac{1}{x} \right) is continuous at x=0x=0
First we have to know about the definition of continuity. Definition of continuity is shown below.
Definition of continuity: ff is continuous at aa , if and only if limxaf(x)=f(a)\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)
Irrespective of the path in which xx tends to aa .
If f(a)f\left( a \right) does not exist, then ff is not continuous at aa.
Therefore, the above written condition describes the definition of continuity.
By using the above written definition of continuity, we have to show that xsin(1x)x\sin \left( \dfrac{1}{x} \right) is continuous at x=0x=0.
First of all, let us find out f(a)f\left( a \right) because if f(a)f\left( a \right) exists then only it will be continuous and if f(a)f\left( a \right) will not exists then the given function is not continuous.
From the question given, we have been asked to show xsin(1x)x\sin \left( \dfrac{1}{x} \right) is continuous at x=0x=0.
Therefore, here a=0a=0
Now,
f(a)=asin(1a)\Rightarrow f\left( a \right)=a\sin \left( \dfrac{1}{a} \right)
f(0)=0sin(10)\Rightarrow f\left( 0 \right)=0\sin \left( \dfrac{1}{0} \right)
Therefore, we can clearly observe that f(a)f\left( a \right) is equal to zero.
Now we need to show that the expression limx0xsin1x=0\displaystyle \lim_{x \to 0}x\sin \dfrac{1}{x}=0 is valid for doing that let us assume 1x=z\dfrac{1}{x}=z then we can write this expression simply as limz1zsinz=0\displaystyle \lim_{z\to \infty }\dfrac{1}{z}\sin z=0 .
From the basic concepts as we know that limθsinθθ=0\displaystyle \lim_{\theta \to \infty }\dfrac{\sin \theta }{\theta }=0 we will use that formulae here.
After using that formula here we will have limz1zsinz=0\displaystyle \lim_{z\to \infty }\dfrac{1}{z}\sin z=0

Therefore, xsin(1x)x\sin \left( \dfrac{1}{x} \right) is continuous at x=0x=0.

Note: We should be well known about the definition of continuity. We should be very careful while checking the continuity. Also, we should be very careful while checking f(a)f\left( a \right). Also, we should be well aware of the usage of the definition of continuity. Also, we should be very careful while writing the definition of continuity because the whole problem is dependent on the definition of continuity. Similarly we can verify the continuity of xcos(1x)x\cos \left( \dfrac{1}{x} \right) or any other function.