Find the following limit- limn - - x2 - 4x - 3-x2 - 3x - 4-

lim_x→∞x^2 4x+3x^2+3x 4=lim_x→∞(x 3)(x 1)(x+4)(x 1)=lim_x→∞x 3x+4=lim_x→∞1 3x1+4x=1 ...

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Standard IX

Mathematics

Properties of Measures of Central Tendency

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Question

Find the following limit.
$lim n \to \infty \frac{x^{2} - 4 x + 3}{x^{2} + 3 x - 4} .$

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Solution

$lim x \to \infty \frac{x^{2} - 4 x + 3}{x^{2} + 3 x - 4} = lim x \to \infty \frac{(x - 3) (x - 1)}{(x + 4) (x - 1)} = lim x \to \infty \frac{x - 3}{x + 4} = lim x \to \infty \frac{1 - \frac{3}{x}}{1 + \frac{4}{x}} = 1$

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Find the following limit.limn→∞x2−4x+3x2+3x−4.

limx→∞x2−4x+3x2+3x−4=limx→∞(x−3)(x−1)(x+4)(x−1)=limx→∞x−3x+4=limx→∞1−3x1+4x=1

Find the following limit.
$lim n \to \infty \frac{x^{2} - 4 x + 3}{x^{2} + 3 x - 4} .$

$lim x \to \infty \frac{x^{2} - 4 x + 3}{x^{2} + 3 x - 4} = lim x \to \infty \frac{(x - 3) (x - 1)}{(x + 4) (x - 1)} = lim x \to \infty \frac{x - 3}{x + 4} = lim x \to \infty \frac{1 - \frac{3}{x}}{1 + \frac{4}{x}} = 1$