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Question: What is \(\underset{x\to 0}{\mathop{\lim }}\,{{\left| x(x-1) \right|}^{\left[ \cos 2x \right]}}\) eq...

What is lim⁡x→0 ∣x(x−1)∣[cos⁡2x]\underset{x\to 0}{\mathop{\lim }}\,{{\left| x(x-1) \right|}^{\left[ \cos 2x \right]}} equal to, where [.]\left[ . \right] denotes greatest integer function?
(a) 1
(b) 0
(c) e
(d) Does not exist

Explanation

Solution

Hint: To evaluate the given limit, simplify the function using the formulas related to the greatest integer function and then apply the limits to the given function to get the value of the limit.

Complete step by step answer:
We have the function of the form ∣x(x−1)∣[cos⁡2x]{{\left| x(x-1) \right|}^{\left[ \cos 2x \right]}}. We have to evaluate the limit of the function around the point x=0x=0.
We know that the function []\left[ {} \right] stands for greatest integer function. This function rounds down a real number to the nearest integer. This function gives the value of the greatest integer that is less than or equal to the given real number.

We also know that lim⁡x→a− [x]=a−1\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\left[ x \right]=a-1 and lim⁡x→a+ [x]=a\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\left[ x \right]=a.
We will evaluate the right hand and left hand limit of the given function.

Firstly, we will evaluate the left hand limit. Thus, we have lim⁡x→0− ∣x(x−1)∣[cos⁡2x]=∣0(−1)∣−1=∞\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{\left| x(x-1) \right|}^{\left[ \cos 2x \right]}}={{\left| 0\left( -1 \right) \right|}^{-1}}=\infty .
On evaluating the right hand limit, we have lim⁡x→0+ ∣x(x−1)∣[cos⁡2x]=∣0(−1)∣0=∞\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{\left| x(x-1) \right|}^{\left[ \cos 2x \right]}}={{\left| 0\left( -1 \right) \right|}^{0}}=\infty .
So, we will use the formula \underset{x\to a}{\mathop{\lim }}\,{{\left\\{ 1+f\left( x \right) \right\\}}^{\dfrac{1}{g\left( x \right)}}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}}} when \underset{x\to a}{\mathop{\lim }}\,{{\left\\{ 1+f\left( x \right) \right\\}}^{\dfrac{1}{g\left( x \right)}}}={{0}^{0}}.

Thus, we have \underset{x\to 0}{\mathop{\lim }}\,{{\left| x(x-1) \right|}^{\left[ \cos 2x \right]}}=\underset{x\to 0}{\mathop{\lim }}\,{{\left\\{ \left| 1+\left( {{x}^{2}}-x-1 \right) \right| \right\\}}^{\left[ \cos 2x \right]}} by factorizing the term ∣x(x−1)∣\left| x\left( x-1 \right) \right|.

So, we have \underset{x\to 0}{\mathop{\lim }}\,{{\left\\{ \left| 1+\left( {{x}^{2}}-x-1 \right) \right| \right\\}}^{\left[ \cos 2x \right]}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left| {{x}^{2}}-x-1 \right|}{{{\left[ \cos 2x \right]}^{-1}}}}}.

Thus, we get {{e}^{\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left| {{x}^{2}}-x-1 \right|}{{{\left[ \cos 2x \right]}^{-1}}}}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\left\\{ \left| -1 \right|\times \cos 0 \right\\}}}=e

Hence, the value of the limit is \underset{x\to 0}{\mathop{\lim }}\,{{\left\\{ \left| 1+\left( {{x}^{2}}-x-1 \right) \right| \right\\}}^{\left[ \cos 2x \right]}}=e, which is option (c).

Note: One must evaluate both the left and right hand side of the limit of the given function. We can’t solve this question without using the formula for calculating the limit of function of the form 00{{0}^{0}}. We should not directly substitute the value of the limit in the function. Otherwise, we will get an incorrect answer.