Solutions

a)

Suppose for contradiction that

$$\lim_{x \to 0} \frac{1}{x}=l.$$

Then, by definition,

$$\forall \varepsilon>0,\ \exists \delta>0,\ 0<|x|<\delta \Rightarrow \left|\frac{1}{x}-l\right|<\varepsilon.$$

But by the reverse triangle inequality, this means that

$$\left|\frac{1}{x}\right|-|l|\leq \left|\frac{1}{x}-l\right|<\varepsilon \Rightarrow \frac{1}{\varepsilon+|l|}<|x|.$$

This in turn means that the inequality does not hold for all $x$ such that $0<|x-a|<\delta$, only for $x$ such that

$$0<\frac{1}{\varepsilon+|l|}<|x|<\delta.$$

Hence, we end up with a contradiction.

b)

We prove this again by contradiction. Suppose that

$$\lim_{x\to 1}\frac{1}{x-1}=l.$$

Then,

$$\forall \varepsilon>0,\ \exists \delta>0,$$$$\begin{align*} 0<|x-1|<\delta &\Rightarrow \left|\frac{1}{x-1}\right|-|l|\leq \left|\frac{1}{x-1}-l\right|<\varepsilon \\ &\Rightarrow \frac{1}{\varepsilon+|l|}<|x-1|. \end{align*}$$

Here again, a contradiction arises from the fact that not all $x$ such that $0<|x-1|<\delta$ guarantee the implied inequality, only $x$ such that

$$0<\frac{1}{\varepsilon+|l|}<|x-1|<\delta.$$