
How do you find \[\displaystyle \lim_{x \to 0}x\cos \left( \dfrac{1}{x} \right)\]?
Answer
570.3k+ views
Hint: Limits of bounded functions are the lower and upper boundaries of the range of the functions. We can split a limit as the product of two limits provided both exist and both are finite, or both are infinite, or one is infinite and the other is finite and nonzero. If the limit of one is zero then the total value would be zero.
Complete step by step answer:
As per the given question, we have to the limit of the given expression at the given limiting condition. Here, the given limiting condition is when x tends to zero and the given expression is \[x\cos \left( \dfrac{1}{x} \right)\]. That is, we have the limit, \[\displaystyle \lim_{x \to 0}x\cos \left( \dfrac{1}{x} \right)\].
We know that the second part of the given expression \[\cos \left( \dfrac{1}{x} \right)\], is a bounded function. For whatever value of \[\left( \dfrac{1}{x} \right)\], the \[\cos \left( \dfrac{1}{x} \right)\] strictly lies between -1 and 1 only. That is, we can write it as
\[\Rightarrow -1<\displaystyle \lim_{x \to 0}\cos \left( \dfrac{1}{x} \right)<1\] ------(1)
Now, let us analyse the first part of the given expression \[x\cos \left( \dfrac{1}{x} \right)\] which x. We know that the limit value of x when x tends to zero is zero itself. That is, this can written as
\[\Rightarrow \displaystyle \lim_{x \to 0}x=0\] -----(2)
Actually, when we multiply anything with zero, we get the same value zero itself.
Therefore, from equation (1) and equation (2), the limit of a bounded function when multiplied by a function whose limit value when x tends to zero is zero must be zero. That is, we can write this as the following equation:
\[\Rightarrow \displaystyle \lim_{x \to 0}x\cos \left( \dfrac{1}{x} \right)=0\]
\[\therefore \] 0 is the limit of \[x\cos \left( \dfrac{1}{x} \right)\] when x tends to 0.
Note:
In order to solve such types of problems, we should have enough knowledge on limits of bounded functions. Like, \[\cos \left( \dfrac{1}{x} \right)\] lies between -1 and 1. We can simply find the value of limit by substituting the limiting of x, here x is 0. And by the fact that multiplication of anything with zero gives zero, we can say that the limit value is zero.
Complete step by step answer:
As per the given question, we have to the limit of the given expression at the given limiting condition. Here, the given limiting condition is when x tends to zero and the given expression is \[x\cos \left( \dfrac{1}{x} \right)\]. That is, we have the limit, \[\displaystyle \lim_{x \to 0}x\cos \left( \dfrac{1}{x} \right)\].
We know that the second part of the given expression \[\cos \left( \dfrac{1}{x} \right)\], is a bounded function. For whatever value of \[\left( \dfrac{1}{x} \right)\], the \[\cos \left( \dfrac{1}{x} \right)\] strictly lies between -1 and 1 only. That is, we can write it as
\[\Rightarrow -1<\displaystyle \lim_{x \to 0}\cos \left( \dfrac{1}{x} \right)<1\] ------(1)
Now, let us analyse the first part of the given expression \[x\cos \left( \dfrac{1}{x} \right)\] which x. We know that the limit value of x when x tends to zero is zero itself. That is, this can written as
\[\Rightarrow \displaystyle \lim_{x \to 0}x=0\] -----(2)
Actually, when we multiply anything with zero, we get the same value zero itself.
Therefore, from equation (1) and equation (2), the limit of a bounded function when multiplied by a function whose limit value when x tends to zero is zero must be zero. That is, we can write this as the following equation:
\[\Rightarrow \displaystyle \lim_{x \to 0}x\cos \left( \dfrac{1}{x} \right)=0\]
\[\therefore \] 0 is the limit of \[x\cos \left( \dfrac{1}{x} \right)\] when x tends to 0.
Note:
In order to solve such types of problems, we should have enough knowledge on limits of bounded functions. Like, \[\cos \left( \dfrac{1}{x} \right)\] lies between -1 and 1. We can simply find the value of limit by substituting the limiting of x, here x is 0. And by the fact that multiplication of anything with zero gives zero, we can say that the limit value is zero.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Giving reasons state the signs positive or negative class 12 physics CBSE

Explain esterification reaction with the help of a class 12 chemistry CBSE

What is defined as a solenoid Depict a diagram with class 12 physics CBSE

