Limit – 1 power infinity – JEE Main 2025 Shift 2

Solution:

The expression can be rewritten as:
$$L = \lim_{x\rightarrow \infty} \left( \frac{2x^2-3x+5}{3x^2+5x+4} \right) \cdot \frac{(3x-1)^{x/2}}{(3x+2)^{x/2}}$$
$$L = \lim_{x\rightarrow \infty} \left( \frac{2 – \frac{3}{x} + \frac{5}{x^2}}{3 + \frac{5}{x} + \frac{4}{x^2}} \right) \cdot \left( \frac{3x-1}{3x+2} \right)^{x/2}$$
The limit of the algebraic part is $\frac{2}{3}$.

Now consider the exponential part:
$$L_{exp} = \lim_{x\rightarrow \infty} \left( \frac{3x(1 – \frac{1}{3x})}{3x(1 + \frac{2}{3x})} \right)^{x/2} = \frac{\lim_{x\rightarrow \infty} (1 – \frac{1}{3x})^{x/2}}{\lim_{x\rightarrow \infty} (1 + \frac{2}{3x})^{x/2}}$$
Using $\lim_{x\to \infty} (1 + \frac{a}{x})^x = e^a$:
Numerator: $\lim_{x\rightarrow \infty} \left( (1 – \frac{1}{3x})^{3x} \right)^{\frac{1}{3x} \cdot \frac{x}{2}} = e^{-1/6}$
Denominator: $\lim_{x\rightarrow \infty} \left( (1 + \frac{2}{3x})^{3x} \right)^{\frac{1}{3x} \cdot \frac{x}{2}} = e^{1/3}$

Combining the limits:
$$L = \frac{2}{3} \cdot \frac{e^{-1/6}}{e^{1/3}} = \frac{2}{3} e^{-1/6 – 1/3} = \frac{2}{3} e^{-1/2}$$
$$L = \frac{2}{3\sqrt{e}}$$

Ans. (4)