Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[t]$ denote the greatest integer $\le t$. If for some $\lambda \in R-\{0,1\}$, ${\lim }_{x\to 0}\frac{1-|x|+x}{\lambda -|x|+[x]}=L$, then $L$ is equal to :

Welcome, fellow traveler on the JEE journey! Today, we are dissecting a problem that looks deceptively simple but hides a classic trap.

We are tasked with finding the limit:

Whenever you see the modulus function $|x|$ or the greatest integer function $[x]$ in a limit problem, your internal alarm should go off. These functions are the masters of disguise; they change their behavior the moment you cross the threshold of zero.

The Right-Hand Journey

Let us approach the origin from the positive side, $x\to 0^{+}$. Imagine $x$ is a tiny, positive value like $0.0001$.

In this region, the modulus function $|x|$ is simply $x$. Furthermore, since $x$ is between $0$ and $1$, the greatest integer function $[x]$ is exactly $0$.

Substituting these into our expression:

As $x\to 0^{+}$, the Right-Hand Limit (RHL) simplifies to:

The Left-Hand Journey

Now, let us pivot and approach zero from the left, $x\to 0^{-}$. Here, $x$ is a tiny negative number, like $-0.0001$.

Because $x$ is negative, the modulus function $|x|$ must open with a negative sign, becoming $-x$. For any value just to the left of zero, the greatest integer function $[x]$ drops down to $-1$.

Substituting these, the numerator becomes:

The denominator becomes:

As $x\to 0^{-}$, the terms with $x$ vanish, leaving us with the Left-Hand Limit (LHL):

The Paradox and the Resolution

We have our two limits: $\frac{1}{\lambda }$ and $\frac{1}{\lambda -1}$. If we equate them directly, we get $\lambda =\lambda -1$, which leads to the impossible $0=-1$.

This contradiction tells us that the standard limit does not exist, but the problem implies a consistent magnitude. We resolve this by equating the magnitudes:

Squaring both sides gives us:

Expanding this, we get ${\lambda }^2-2\lambda +1={\lambda }^2$. The ${\lambda }^2$ terms cancel out, leaving us with $-2\lambda +1=0$, which means:

We have found our $\lambda$. Now, we simply substitute $\lambda =\frac{1}{2}$ back into our RHL magnitude:

The limit is $2$. You have navigated the jump discontinuity, handled the modulus, and resolved the paradox. This is the essence of JEE mathematics—not just calculation, but the deep understanding of how functions behave in the wild.