JAMIA 2026 — Mathematics PYQ

Step 1: Divide the numerator and denominator by $\sqrt{x}$

Let's divide both top and bottom by $\sqrt{x}$:

$$L=\underset{x\to \infty }{\lim }\frac{\frac{\sqrt{x}}{\sqrt{x}}}{\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x}}}$$

The numerator simplifies completely to $1$:

$$L=\underset{x\to \infty }{\lim }\frac{1}{\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x}}}$$

Step 2: Simplify the denominator

Using the algebraic property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$, we can bring the $\sqrt{x}$ inside the main square root of the denominator:

$$\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x}}=\sqrt{\frac{x+\sqrt{x+\sqrt{x}}}{x}}$$

Now, divide each term inside the main square root by $x$:

$$=\sqrt{\frac{x}{x}+\frac{\sqrt{x+\sqrt{x}}}{x}}$$

$$=\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}}$$

To pull the remaining $x$ in the denominator into the nested square root, rewrite $x$ as $\sqrt{x^2}$:

$$=\sqrt{1+\sqrt{\frac{x+\sqrt{x}}{x^2}}}$$

$$=\sqrt{1+\sqrt{\frac{x}{x^2}+\frac{\sqrt{x}}{x^2}}}$$

$$=\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{x}{x^4}}}}$$

$$=\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{3/2}}}}$$

Step 3: Substitute the limit $x\to \infty$

Now put the simplified denominator back into our original limit expression:

$$L=\underset{x\to \infty }{\lim }\frac{1}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x^{3/2}}}}}$$

As $x\to \infty$, any term with $x$ in its denominator approaches $0$ (i.e., $\frac{1}{x}\to 0$ and $\frac{1}{x^{3/2}}\to 0$):

$$L=\frac{1}{\sqrt{1+\sqrt{0+0}}}$$

$$L=\frac{1}{\sqrt{1+0}}$$

$$L=\frac{1}{1}=1$$

Correct Answer:

(b) 1