lim _{x rightarrow infty} frac{(2x^2-3x+5)(3x | vedclass

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Question

(D) दिया गया सीमा: $L = \lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)(3x+2)^{x/2}}$$= \lim _{x \rightarrow \infty} \frac{x^2

Vedclass · Accepted Answer

$\frac{2}{3\sqrt{e}}$

Vedclass · Answer

(D) दिया गया सीमा: $L = \lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)(3x+2)^{x/2}}$$= \lim _{x \rightarrow \infty} \frac{x^2(2-3/x+5/x^2)}{x^2(3+5/x+4/x^2)} \cdot \left( \frac{3x-1}{3x+2} \right)^{x/2}$$= \frac{2}{3} \cdot \lim _{x \rightarrow \infty} \left( \frac{1-1/(3x)}{1+2/(3x)} \right)^{x/2}$$= \frac{2}{3} \cdot \frac{\lim _{x \rightarrow \infty} (1-1/(3x))^{x/2}}{\lim _{x \rightarrow \infty} (1+2/(3x))^{x/2}}$$= \frac{2}{3} \cdot \frac{e^{\lim _{x \rightarrow \infty} (x/2)(-1/(3x))}}{e^{\lim _{x \rightarrow \infty} (x/2)(2/(3x)))}}$$= \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} = \frac{2}{3\sqrt{e}}$

$\lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}}$ का मान ज्ञात कीजिए।

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