Question

Let \[\left[ t \right]\] denote the greatest integer \[ \le t\]. If for some \[\lambda \in R - \left\{ {0,1} \right\}\] , \[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - x + \left| x \right|}}{{\lambda - x + \left[ x \right]}} = L\]. Then find the value of \[L\].
A. 0
B. 2
C. \[\dfrac{1}{2}\]
D. 1

Answer 1

Hint: In the given question the equation of a limit is given . First, calculate the left-hand limit and right-hand limit. Then equate the limits and find the value of \[L\].

Formula used:
Left hand limit: The limit of a function \[f\left( x \right)\], as \[x\] approaches to \[a\] from the left is, \[\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L\]
Right hand limit: The limit of a function \[f\left( x \right)\], as \[x\] approaches to \[a\] from the right is, \[\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = L\]

Complete step by step solution:
The given equation of limit is \[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - x + \left| x \right|}}{{\lambda - x + \left[ x \right]}} = L\], where \[\lambda \in R - \left\{ {0,1} \right\}\].
The left-hand limit of the given equation is,
\[LHL = \mathop {lim}\limits_{x \to {0^ - }} \left| {\dfrac{{1 - x + \left| x \right|}}{{\lambda - x + \left[ x \right]}}} \right|\]
\[ \Rightarrow \]\[LHL = \mathop {lim}\limits_{h \to {0^ - }} \left| {\dfrac{{1 - h - h}}{{\lambda - h - 1}}} \right|\]
\[ \Rightarrow \]\[LHL = \left| {\dfrac{{1 - 0 - 0}}{{\lambda - 0 - 1}}} \right|\]
\[ \Rightarrow \]\[LHL = \left| {\dfrac{1}{{\lambda - 1}}} \right|\]

The right-hand limit of the given equation is,
\[RHL = \mathop {lim}\limits_{x \to {0^ + }} \left| {\dfrac{{1 - x + \left| x \right|}}{{\lambda - x + \left[ x \right]}}} \right|\]
\[ \Rightarrow \]\[RHL = \mathop {lim}\limits_{h \to {0^ + }} \left| {\dfrac{{1 - h + h}}{{\lambda - h + 0}}} \right|\]
\[ \Rightarrow \]\[RHL = \left| {\dfrac{1}{{\lambda - 0}}} \right|\]
\[ \Rightarrow \]\[RHL = \left| {\dfrac{1}{\lambda }} \right|\]
For existence of a limit, \[RHL = LHL\].
Then,
    \[\left| {\dfrac{1}{\lambda }} \right| = \left| {\dfrac{1}{{\lambda - 1}}} \right|\]
\[ \Rightarrow \]\[\left| \lambda \right| = \left| {\lambda - 1} \right|\]
Take square on both sides.
\[{\lambda ^2} = {\lambda ^2} - 2\lambda + 1\]
Simplify the above equation.
\[ \Rightarrow \]\[2\lambda = 1\]
\[ \Rightarrow \]\[\lambda = \dfrac{1}{2}\]

Substitute \[\lambda = \dfrac{1}{2}\] in the given equation of a limit.
\[L = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - x + \left| x \right|}}{{\dfrac{1}{2} - x + \left[ x \right]}}\]
\[ \Rightarrow \]\[L = \dfrac{{1 - 0 + \left| 0 \right|}}{{\dfrac{1}{2} - 0 + \left[ 0 \right]}}\]
\[ \Rightarrow \]\[L = \dfrac{1}{{\dfrac{1}{2}}}\]
\[ \Rightarrow \]\[L = 2\]
Hence the correct option is option B.

Note: Students are often confused with the modulus function. The mathematical representation of the modulus function is, \[f\left( x \right) = \left| x \right| = \left\{ \begin{array}{l}x, if x \ge 0\\ - x, if x < 0\end{array} \right.\].
If the value of \[x\] is less than zero, then the output is minus of the original value. If the value of \[x\] is greater than or equal to zero, then the output is the original value.