Solutions

a)

In the forward $(\Rightarrow)$ implication, suppose

$$\lim_{x \to a} f(x)=l.$$

Let $g(x)=-l$. It is trivially true that

$$\lim_{x \to a} g(x) = -l.$$

By the theorem on limits of sums, we have that

$$\lim_{x \to a}[f(x)+g(x)] = l-l = 0.$$

Conversely $(\Leftarrow)$, suppose

$$\lim_{x \to a}[f(x)-l]=0.$$

By definition,

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|x-a|<\delta \implies |(f(x)-l)-0|<\epsilon$$$$\implies |f(x)-l|<\epsilon,$$

which is also the definition for $\lim_{x \to a} f(x)=l$.

b)

Let $\lim_{x \to 0} f(x)=l$. By definition,

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|x|<\delta \implies |f(x)-l|<\epsilon.$$

Let $z=x-a$. Then

$$0<|z|<\delta \equiv 0<|x-a|<\delta \implies |f(z)-l|<\epsilon,$$$$\implies |f(x-a)-l|<\epsilon,$$

which implies that $\lim_{x \to a} f(x-a)=l$ also.

Conversely $(\Leftarrow)$, suppose $\lim_{x \to a} f(x-a)=l$. By definition,

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|x-a|<\delta \implies |f(x-a)-l|<\epsilon.$$

Again, let $z=x-a$. Then,

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|z|<\delta \implies |f(z)-l|<\epsilon,$$

which means $\lim_{z \to 0} f(z)=l$.

c)

Let $\lim_{x \to 0} f(x)=l$. By definition,

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|x|<\delta \implies |f(x)-l|<\epsilon.$$

Let $\delta = \min(1,\delta_1)$. Then,

$$0<|x|<\delta \implies 0<|x^3|<\delta \implies |f(x^3)-l|<\epsilon,$$

which means $\lim_{x \to 0} f(x^3)=l$.

Conversely $(\Leftarrow)$, let

$$\lim_{x \to 0} f(x^3)=l.$$

By definition,

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|x^3|<\delta \implies |f(x^3)-l|<\epsilon.$$

Since every number has a cube root, let $z=x^3$. Then, the above can be rewritten as

$$\forall \epsilon > 0,\ \exists \delta > 0,\ 0<|z|<\delta \implies |f(z)-l|<\epsilon,$$

which means $\lim_{z \to 0} f(z)=l$.

d)

Let

$$f(x)= \begin{cases} 1, & x \ge 0, \\ 0, & \text{otherwise}. \end{cases}$$

We have

$$\lim_{x \to 0} f(x^2)=1,$$

but $\lim_{x \to 0} f(x)$ does not exist.