Animated Solution for Mathematics - Limits, Continuity and Differentiability: For each $t\in \mathbb{R}$, let $[t]$ be the greatest integer less than or equal to $t$. Then ${\lim }_{x\to 0^{+}}x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\cdots +\left[\frac{15}{x}\right]\right)$

Welcome, fellow traveler on the JEE Advanced journey. Today, we confront a problem that often strikes fear into the hearts of students: the Greatest Integer Function (GIF), denoted by $[t]$.

When you see a limit like ${\lim }_{x\to 0^{+}}x{\sum }_{k=1}^{15}[\frac{k}{x}]$, it is natural to feel a moment of hesitation. We are staring at a sum of fifteen different GIF terms, all being squeezed by an $x$ that is vanishing into zero.

It looks like a classic $0\cdot \infty$ indeterminate form, doesn't it? But fear not. We are not here to guess; we are here to dismantle the problem with surgical precision.

To solve this, we need a tool that separates the 'integer' part from the 'messy' fractional part. That tool is the fundamental identity: $[t]=t-\{t\}$.

Here, $\{t\}$ is the fractional part of $t$, which, by definition, is always trapped in the interval $[0,1)$. Think of this as a way to 'linearize' the GIF.

Instead of dealing with the jagged, step-like nature of the GIF, we are now dealing with a smooth linear term $t$ and a bounded, oscillating term $\{t\}$.

Let's apply this to every single term in our summation. For any $k$, we write:

When we substitute this into our original limit, the expression transforms into:

Now, watch the magic of distribution. We can split this large summation into two distinct groups:

Look at the first group. The $x$ outside the bracket perfectly cancels the $x$ in the denominator of each term $\frac{k}{x}$. The $x$ vanishes entirely!

We are left with the sum of the first 15 natural numbers: $1+2+3+\cdots +15$. Using the arithmetic progression formula $\frac{n(n+1)}{2}$, where $n=15$, we calculate:

The first part of our limit is a solid, unshakeable constant: $120$.

Now, what about the second group? We have $x$ multiplied by the sum of fifteen fractional parts: $x{\sum }_{k=1}^{15}\{\frac{k}{x}\}$.

We know that for any real number $t$, the fractional part $\{t\}$ is always between $0$ and $1$. Therefore, the sum of fifteen such terms must be bounded:

We are taking the limit as $x\to 0^{+}$. We have a term $x$ that is shrinking to zero, multiplied by a sum that is trapped between $0$ and $15$.

By the Squeeze Theorem, or simply by the logic that zero times any finite, bounded value is zero, this entire second term collapses into nothingness. It vanishes!

We are left with $120-0=120$. The complexity of the GIF has been stripped away, leaving behind a beautiful, clean integer.

This is the elegance of JEE Advanced mathematics—taking a seemingly chaotic, indeterminate expression and finding the underlying order.

Remember this technique; it is a powerful weapon in your arsenal for any limit problem involving the Greatest Integer Function. The final answer is 120.