Math, asked by tkmuthuashif44, 1 year ago

lim x->infinity 5x^2+8x-3/3x^2+2

Answers

Answered by rational
2
Since the degree of numerator and denominator is same(2), the limit equals the ratio of leading coefficients : \dfrac{5}{3}
Answered by kvnmurty
2
Let\ v=\frac{1}{x},\ \ as\ x \to\ \infty,\ v \to\ 0.\\\\ \lim_{x \to \infty} \frac{5x^2 + 8 x - 3}{3x^2+2}\\\\= \lim_{v \to 0} \frac{\frac{5x^2+8x-3}{x^2}}{\frac{3x^2+2}{x^2}}\\\\=\lim_{v \to 0} \frac{5+\frac{8}{x}-\frac{3}{x^2}}{3+\frac{2}{x^2}}\\\\=\lim_{v \to 0} \frac{5+8v-3v^2}{3+2v^2}\\\\=\lim_{v \to 0} \frac{5+8*0-3*0}{3+2*0}\\\\=\frac{5}{3}
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