Answer
(a) $\lim_{x\to0}f(x)=0$
(b) $\lim_{x\to0}\frac{f(x)}{x}=0$
Work Step by Step
$$\lim_{x\to0}\frac{f(x)}{x^2}=1$$
(a) Here we cannot use the substitution method like normal, since if we use that method, the denominator would end up with $0$.
Instead, we can rewrite $f(x)$ as follows:
$$f(x)=\frac{f(x)}{x^2}\times(x^2)$$
Therefore, $$\lim_{x\to0}f(x)=\lim_{x\to0}\Big(\frac{f(x)}{x^2}\times(x^2)\Big)$$ $$\lim_{x\to0}f(x)=\lim_{x\to0}\frac{f(x)}{x^2}\times\lim_{x\to0}(x^2)$$ $$\lim_{x\to0}f(x)=1\times0^2=0$$
(b) Again, we cannot use the substitution method here, but we would rewrite: $$\frac{f(x)}{x}=\frac{f(x)}{x^2}\times x$$
Therefore, $$\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\Big(\frac{f(x)}{x^2}\times(x)\Big)$$ $$\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{f(x)}{x^2}\times\lim_{x\to0}(x)$$ $$\lim_{x\to0}\frac{f(x)}{x}=1\times0=0$$
