I deduced this proof from the proof of the limit of a reciprocal property (Michael Spivak Calculus 3rd edition, pages 103-104). The exercise goes as follows:
Exercise. Using epsilon-delta definition, show that $\displaystyle\lim_{x \to 1} \left[f(x) = \dfrac{100}{x}\right] = 100$.
Lemma. If $y_0 \not = 0$ and $$|y-y_0| < \min\left(\dfrac{|y_0|}{2}, \dfrac{\epsilon|y_0|^2}{2}\right)$$ then $y \not = 0$ and $$\left| \dfrac{1}{y}-\frac{1}{y_0}\right|< \epsilon$$.
This lemma is true and its proof can be found in the aforementioned text, pages 102-103.
Proof
Let $g(x) = \dfrac{x}{100}$. Then
\begin{align} \lim_{x \to 1} g(x)&= \lim_{x \to 1}x \cdot \lim_{x \to 1} \dfrac{1}{100} && \text{this by product property of limits}\\ &= \frac{1}{100} \end{align} $$$$
From this we have that for all $\epsilon_{1}>0$, there exists $\delta_1>0$ such that, for all $x$ if $$0 < |x-1| < \delta_1$$ then $$\left|\dfrac{x}{100} - \dfrac{1}{100}\right|< \epsilon_1$$ Because $$\min\left(\dfrac{\left|\frac{1}{100}\right|}{2}, \dfrac{\epsilon\left(\left|\frac{1}{100}\right|\right)^2}{2}\right) > 0$$ then if $\epsilon > 0$ there exists $\delta > 0$ such that, for all $x$ if $$0 < |x-1| < \delta $$ then $$\left|g(x) - \dfrac{1}{100}\right| = \left|\dfrac{x}{100} - \dfrac{1}{100}\right|< \min\left(\dfrac{\left|\frac{1}{100}\right|}{2}, \dfrac{\epsilon\left(\left|\frac{1}{100}\right|\right)^2}{2}\right) $$ By lemma, we therefore have that $g(x) \not = 0$ and $$\left|\dfrac{1}{g(x)} - \dfrac{1}{\frac{1}{100}}\right| = \left|\dfrac{100}{x} - 100\right| < \epsilon$$.
Thanks for your attention.