lim _{x rightarrow-infty} frac{5 x^3-x^2 sin | vedclass

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Question

(C) આપેલ લક્ષ $L = \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3}$ છે.જ્યારે $x \rightarrow -\infty$,ત્યારે $x < 0$

Vedclass · Accepted Answer

$-5/7$

Vedclass · Answer

(C) આપેલ લક્ષ $L = \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3}$ છે.જ્યારે $x \rightarrow -\infty$,ત્યારે $x આ કિંમતો પદાવલિમાં મૂકતા:$L = \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x + 7(-x^3) - 4(-x) + 3} = \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x - 7x^3 + 4x + 3}$.અંશ અને છેદને $x^3$ વડે ભાગતા:$L = \lim _{x \rightarrow-\infty} \frac{5 - \frac{\sin 5 x}{x}}{\frac{\cos 4 x}{x^2} - 7 + \frac{4}{x^2} + \frac{3}{x^3}}$.જેમ $x \rightarrow -\infty$,તેમ $\frac{\sin 5 x}{x} \rightarrow 0$,$\frac{\cos 4 x}{x^2} \rightarrow 0$,$\frac{4}{x^2} \rightarrow 0$,અને $\frac{3}{x^3} \rightarrow 0$.તેથી,$L = \frac{5 - 0}{0 - 7 + 0 + 0} = -\frac{5}{7}$.

$\lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3} = $

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