Find the value ofa.) lim (x²-1)/(x²+1)x−>∞

Mathematics

1 answer:

Fed [463]4 years ago

6 0

Answer:

$\displaystyle\lim_{x \to \infty} \frac{x^2-1}{x^2+1} \\ \text{take $x^2$ common from numerator and denominator} \\ \displaystyle\lim_{x \to \infty} \dfrac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}} \\\text{Now put the limit}\\\displaystyle\lim_{x \to \infty} \frac{1-0}{1+0}= 1$

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The equation $48a^3-75a$ is factorable using GCF and difference of squares.

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The term in the parenthesis ${16a^2-25)$ has the form $a^2-b^2 = (a-b)(a+b)$ which factors into two terms where a and b are the squares root of their respective terms.