Answer to Question #213707 in Calculus for amit

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)

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

does not exist.

$\lim\limits_{y\to 1}(\sqrt[4]{4y^3}-3)=\sqrt{2}-3$

(b)

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

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

(c)

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

$=-\dfrac{1}{2+2}=-\dfrac{1}{4}$

(d)

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

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

$\big[\dfrac{0}{0}\big]$

L'Hospital's Rule

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

$=\lim\limits_{x\to 0}\dfrac{\cos x-\cos x+x\sin x}{3x^2}=\lim\limits_{x\to 0}\dfrac{\sin x}{3x}$

$=\lim\limits_{x\to 0}\dfrac{(\sin x)'}{(3x)'}=\lim\limits_{x\to 0}\dfrac{\cos x}{3}=\dfrac{1}{3}$