Answer
$$\lim_{x\to0^+}\Big(\frac{1}{x^{1/3}}+\frac{1}{(x-1)^{4/3}}\Big)=\infty$$
$$\lim_{x\to0^-}\Big(\frac{1}{x^{1/3}}+\frac{1}{(x-1)^{4/3}}\Big)=-\infty$$
$$\lim_{x\to1^+}\Big(\frac{1}{x^{1/3}}+\frac{1}{(x-1)^{4/3}}\Big)=-\infty$$
$$\lim_{x\to1^-}\Big(\frac{1}{x^{1/3}}+\frac{1}{(x-1)^{4/3}}\Big)=-\infty$$
Work Step by Step
(a) $$A=\lim_{x\to0^+}\Big(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}}\Big)=\lim_{x\to0^+}\frac{1}{x^{1/3}}-\frac{1}{(0-1)^{4/3}}$$ $$A=\lim_{x\to0^+}\frac{1}{x^{1/3}}-\frac{1}{(-1)^{4/3}}=\lim_{x\to0^+}\frac{1}{x^{1/3}}-\frac{1}{1}$$ $$A=\lim_{x\to0^+}\frac{1}{x^{1/3}}-1$$
As $x\to0^+$, $x^{1/3}\to0^+\gt0$, so the function will approach $\infty$. Therefore, $$A=\infty-1=\infty$$
(b) $$B=\lim_{x\to0^-}\Big(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}}\Big)=\lim_{x\to0^-}\frac{1}{x^{1/3}}-\frac{1}{(0-1)^{4/3}}$$ $$B=\lim_{x\to0^-}\frac{1}{x^{1/3}}-\frac{1}{(-1)^{4/3}}=\lim_{x\to0^-}\frac{1}{x^{1/3}}-\frac{1}{1}$$ $$B=\lim_{x\to0^-}\frac{1}{x^{1/3}}-1$$
As $x\to0^-$, $x^{1/3}\to0^-\lt0$, so the function will approach $-\infty$. Therefore, $$B=-\infty-1=-\infty$$
(c) $$C=\lim_{x\to1^+}\Big(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}}\Big)=-\lim_{x\to1^+}\frac{1}{(x-1)^{4/3}}+\frac{1}{1^{1/3}}$$ $$C=-\lim_{x\to1^+}\frac{1}{(x-1)^{4/3}}+\frac{1}{1}=-\lim_{x\to1^+}\frac{1}{(x-1)^{4/3}}+1$$
As $x\to1^+$, $(x-1)^{4/3}\to0^+\gt0$, so the function will approach $\infty$. Therefore, $$C=-(\infty)+1=-\infty$$
(d) $$D=\lim_{x\to1^-}\Big(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}}\Big)=-\lim_{x\to1^-}\frac{1}{(x-1)^{4/3}}+\frac{1}{1^{1/3}}$$ $$D=-\lim_{x\to1^-}\frac{1}{(x-1)^{4/3}}+\frac{1}{1}=-\lim_{x\to1^-}\frac{1}{(x-1)^{4/3}}+1$$
As $x\to1^-$, $(x-1)^{4/3}\to0^+\gt0$ (because $(x-1)^{4/3}\ge0$ for all $x$), so the function will approach $\infty$. Therefore, $$D=-(\infty)+1=-\infty$$
