Answer to Question #213706 in Calculus for rahul

Question

Evaluate the following limits

(a) Lim y$\to$ -1 Fourth root of 4y^3 minus 3

(b) lim x$\to$ infinity 1+x all over 1-x

(c) lim x $\to$ 2 x - 2 all over 4 - x^2

(d) lim x $\to$ 0 sinx - x cos x all over x^3

Answer 1

(a) $\lim\limits_{y\to-1}\sqrt[4]{4y^3-3}=?$

If $y<0$ then $4y^3-3<0$ and the fourth root doesn't exist in real numbers.

So we will consider another limit:

$\lim\limits_{y\to 1}\sqrt[4]{4y^3-3}=\lim\limits_{y\to 1}\sqrt[4]{1+4(y^3-1)}=\lim\limits_{y\to 1}\sqrt[4]{1}=1$

(b) $\lim\limits_{x\to\infty}\frac{1+x}{1-x}=\lim\limits_{x\to\infty}\frac{x^{-1}+1}{x^{-1}-1}=\lim\limits_{x\to\infty}\frac{1}{-1}=-1$

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

(d) $\lim\limits_{x\to 0}\frac{\sin x-x\cos x}{x^3}=\lim\limits_{x\to 0}\frac{x-x^3/6+o(x^3)-x(1-x^2/2+o(x^2))}{x^3}=$

$=\lim\limits_{x\to 0}\frac{-x^3/6+x^3/2+o(x^3)}{x^3}=\lim\limits_{x\to 0}(-\frac{1}{6}+\frac{1}{2}+o(1))=\frac{1}{3}$