JEE Main 2025 Jan 23 Shift 2, Mathematics Q20: Methods to Evaluate Limits

JEE Main2025Jan 23, Shift 2Mathematics

$x \to \infty lim \frac{( 2 x ^{2} - 3 x + 5 ) ( 3 x - 1 ) ^{x /2}}{( 3 x ^{2} + 5 x + 4 ) ( 3 x + 2 ) ^{x}}$ is equal to :

Solution

Note $(3 x + 2)^{x} = (3 x + 2)^{x /2}$ . Split the limit into a rational factor and an exponential factor:

$L = x \to \infty lim \frac{2 x ^{2} - 3 x + 5}{3 x ^{2} + 5 x + 4} \cdot (\frac{3 x - 1}{3 x + 2})^{x /2} .$

Rational factor: divide numerator and denominator by $x^{2}$ to get $\frac{2 - 3/ x + 5/ x ^{2}}{3 + 5/ x + 4/ x ^{2}} \to \frac{2}{3} .$

Exponential factor: $\frac{3 x - 1}{3 x + 2} = 1 - \frac{3}{3 x + 2}$ . Take logarithm:

$\frac{x}{2} ln (1 - \frac{3}{3 x + 2}) \sim \frac{x}{2} \cdot (- \frac{3}{3 x + 2}) \to - \frac{1}{2}$ as $x \to \infty.$

So the exponential factor approaches $e^{- 1/2}$ .

$L = \frac{2}{3} \cdot e^{- 1/2} = \frac{2}{3 e} .$

Concept behind this question

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