I'm currently self studying Spivak's calculus, but the problems in the back of each chapter are quite difficult. I'm not sure where to start on this problem.
Find a $\delta$ such that $|f(x)-l| < \epsilon$ for all $x$ satisfying $0 < |x-a| < \delta$ for $f(x) = \frac1x$; $a =1$, $l =1$.
On
Hint. Note that for $x> 1/2$, $$|f(x)-l|=\left|\frac{1}{x}-1\right|=\frac{|x-1|}{|x|}<\frac{|x-1|}{1/2}=2|x-1|$$ Can you take it from here?
On
first you can rewrite $\left|\frac{1}{x}-1\right|$ as $\frac{|x-1|}{|x|}$ and let's say $\delta=\frac{1}{2}$
now: $$|x-1|<\frac{1}{2}\implies-\frac{1}{2}<x-1<\frac{1}{2}\implies \frac{1}{2}<x<\frac{3}{2}\implies \frac{1}{2}<|x|<\frac{3}{2}\\\implies \frac{2}{3}<\frac{1}{|x|}<2$$
now we want to return to the original form:$$\frac{1}{|x|}<2\implies\frac{|x-1|}{|x|}<2|x-1|$$
now lets say that $\delta=\frac{\epsilon}{2}$(try to see why it works yourself), but now we have 2 values for $\delta$, so, we can see that $\delta=\frac{1}{2}$ works for all values of $x$ that are larger than $\frac{1}{2}$, so lets do the following:$$\delta=min\left(\frac{\epsilon}{2},\frac{1}{2}\right)$$
try to do the checking(or The Proof if you want to call it like this) alone.
I'm hoping I'm clear enough, ask if you don't understand something
Note that $$ \left|\frac{1}{x}-1\right|=\left|\frac{1-x}{x}\right|=\frac{|x-1|}{|x|}\tag{1} $$ Suppose that $|x-1|<\frac{1}{2}$. Then $$ 1=|1|=|(1-x)+x|\leq |1-x|+|x|=|x-1|+|x|<\frac{1}{2}+|x| $$ and hence $$ |x|>1-\frac{1}{2}=\frac{1}{2}\Longleftrightarrow \frac{1}{|x|}<2 $$ Going back to $(1)$, we have $$ \left|\frac{1}{x}-1\right|=\frac{|x-1|}{|x|}<2|x-1| $$ We want to make $\left|\frac{1}{x}-1\right|<\epsilon$ which forces $2|x-1|<\epsilon$ and so $|x-1|<\frac{\epsilon}{2}$. Hence, we choose $\delta=\min\left(\frac{1}{2},\frac{\epsilon}{2}\right)$.
Note that there is nothing special about $\frac{1}{2}$. We could use $\frac{1}{3}$ but then we would choose $\delta=\min\left(\frac{1}{3},\frac{2\epsilon}{3}\right)$.