Question

Let $\left[t\right]$ denote the greatest integer$\le t$. If for some $\lambda \in R-\left\{0,1\right\}$, $\underset{x\to 0}{lim}\left|\frac{1-x+\left|x\right|}{\lambda -x+\left[x\right]}\right|=L$ then $L$ is equal to.

• A. $0$
• B. $2$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is B

$2$

Explanation for the correct option:

Finding the value of by evaluating $L$ the limit:

Evaluating $\underset{x\to 0}{lim}\left|\frac{1-x+\left|x\right|}{\lambda -x+\left[x\right]}\right|=L$

Now Right-Hand Limit

$$\begin{gathered} \underset{x\to 0+h}{lim}\left|\frac{1-x+\left|x\right|}{\lambda -x+\left[x\right]}\right| \\ =\underset{h\to 0}{l\mathrm{im}}\left|\frac{1-h+h}{\lambda -h+\left[h\right]}\right| \\ =\left|\frac{1}{\lambda -h+0}\right|\left[\because \left[h\right]=0\text{at}h=0\right] \\ =\left|\frac{1}{\lambda }\right| \\ =\frac{1}{\left|\lambda \right|} \end{gathered}$$

Now Left- Hand Limit

$$\begin{gathered} \underset{x\to 0-h}{lim}\left|\frac{1-x+\left|x\right|}{\lambda -x+\left[x\right]}\right| \\ =\underset{h\to 0}{l\mathrm{im}}\left|\frac{1-\left(-h\right)+\left|-h\right|}{\lambda -(-h)+\left|-h\right|}\right| \\ =\underset{h\to 0}{l\mathrm{im}}\left|\frac{1+2h}{\lambda +2h}\right| \\ \mathrm{Applying}\mathrm{limits} \\ =\left|\frac{1+2\left(0\right)}{\lambda +2\left(0\right)}\right| \\ =\frac{1}{\left|\lambda \right|} \end{gathered}$$

Since Right Hand Limit = Left Hand Limit $=\frac{1}{\left|\lambda \right|}$ ….$\left(i\right)$

Thus a limit exists.

$$\begin{gathered} \because \left|\lambda \right|=|\lambda -1|,[-h]=-1 \\ \Rightarrow {\lambda }^2={\lambda }^2-2\lambda +1\mathbf{[}\mathbf{}\mathbf{\text{squaring both sides}}\mathbf{}\mathbf{]} \\ \Rightarrow \lambda =\frac{1}{2} \end{gathered}$$

From $\left(i\right)$, $L=2$

Therefore, the correct answer is option (B).