Answer to Question #212381 in Calculus for Kenneth

Question

1. 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)

$$\lim\limits_{y\to-1}(\sqrt[4]{y^3}-3)=DNE\ (\text{does not exist)}$$

since we consider $y\in \R, y\geq0.$

(b)

$$\lim\limits_{x\to\infin}(\dfrac{1+x}{1-x})=\lim\limits_{x\to\infin}\bigg(\dfrac{\dfrac{1}{x}+\dfrac{x}{x}}{\dfrac{1}{x}-\dfrac{x}{x}}\bigg)$$

$$=\lim\limits_{x\to\infin}\bigg(\dfrac{\dfrac{1}{x}+1}{\dfrac{1}{x}-1}\bigg)=\dfrac{0+1}{0-1}=-1$$

(c)

$$\lim\limits_{x\to2}(\dfrac{x-2}{4-x^2})=\lim\limits_{x\to2}\bigg(\dfrac{x-2}{(2-x)(2+x)}\bigg)$$

$$=\lim\limits_{x\to2}\big(\dfrac{-1}{2+x}\big)=-\dfrac{1}{2+2}=-\dfrac{1}{4}$$

(d)

$$\lim\limits_{x\to0}(\dfrac{\sin x-x\cos x}{x^3})$$

$$\lim\limits_{x\to0}(\sin x-x\cos x)=\sin(0)-0(\cos(0))=0$$

$$\lim\limits_{x\to0}(x^3)=(0)^3=0$$

$$\dfrac{0}{0}=>\lim\limits_{x\to0}(\dfrac{\sin x-x\cos x}{x^3})=\lim\limits_{x\to0}\dfrac{(\sin x-x\cos x)'}{(x^3)'}$$

$$=\lim\limits_{x\to0}\dfrac{\cos x-\cos x+x\sin x}{3x^2}=\lim\limits_{x\to0}\dfrac{\sin x}{3x}$$

$$=\dfrac{1}{3}\lim\limits_{x\to0}\dfrac{\sin x}{x}=\dfrac{1}{3}(1)=\dfrac{1}{3}$$

We use that $\lim\limits_{x\to0}\dfrac{\sin x}{x}=1$