Is there a neater proof for $\lim_{x \rightarrow 1} 1/x = 1$?

Question

What I need to prove:

For all $\epsilon>0$, there exist a $\delta>0$, such that for all $x$ satisfying $0<|x-1|<\delta$, we have $|1/x-1|<\epsilon$.

My proof:

Let $\delta=\min \{ 1/2 ,\epsilon/2 \}$. Then when $\epsilon\geq 1$, $$ |x-1| < \frac{1}{2} \\ (*) \frac{1}{2} < x < \frac{3}{2} \\ \frac{2}{3} < \frac{1}{x} < 2 \\ -\frac{1}{3} < \ \frac{1}{x}-1 < 1 $$ and since $-1<-1/3$, we have $|1/x-1|<1\leq \epsilon$.

When $\epsilon<1$, $$ |x-1| < \frac{\epsilon}{2} \\ \frac{2-\epsilon}{2} < x < \frac{2+\epsilon}{2} \\ -\frac{\epsilon}{2+\epsilon} < \frac{1}{x}-1 < \frac{\epsilon}{2-\epsilon} $$

and since $\epsilon>0$, we have $2+\epsilon > 2-\epsilon$. So $$ -\frac{\epsilon}{2-\epsilon} < -\frac{\epsilon}{2+\epsilon} \implies \left| \frac{1}{x}-1 \right|< \frac{\epsilon}{2-\epsilon} < \epsilon. $$

What I'm looking for is a proof that doesn't "deconstruct" the absolute values as in the starred inequality, ie uses only absolute value identities such as the triangle inequality and its variations.

Answer 1

If $|x - 1| \leqslant 1/2$, then $1 - x \leqslant |1 - x| = |x - 1| \leqslant 1/2$, therefore $x \geqslant 1/2$.

Since $x > 0$, the expression $1/x$ is well-defined.

Also because $x > 0$, we have $|x| = x \geqslant 1/2$.

If you must use the Triangle Inequality, you can argue (using it in the form $|u - v| \geqslant ||u| - |v||$): $$ |x| = |1 - (1 - x)| \geqslant |1 - |1 - x|| = |1 - |x - 1|| \geqslant 1/2 \quad (\text{because } 1 - |x - 1| \geqslant 1/2 > 0), $$ but that only seems to make the argument more complicated, not less.

(It looks a bit less unnatural, however, if you start with the substitution $x = 1 - h$, or $x = 1 + h$.)

Therefore, if $|x - 1| \leqslant 1/2$, $$ \left\lvert\frac{1}{x} - 1\right\rvert = \left\lvert\frac{1 - x}{x}\right\rvert = \frac{|1 - x|}{|x|} = \frac{|x - 1|}{|x|} \leqslant 2|x - 1|. $$ Therefore, for all $\epsilon > 0$, $$ \text{if } 0 < |x - 1| < \frac{\min\left\{1, \epsilon\right\}}{2}, \text{ then } \left\lvert\frac{1}{x} - 1\right\rvert < \epsilon. $$ So, just as you found, we can take: $$ \delta = \frac{\min\left\{1, \epsilon\right\}}{2}. $$

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If you were asked to prove, more generally, $$ a \ne 0 \implies \lim_{x \to a} \frac{1}{x} = \frac{1}{a} $$ (directly, that is, rather than deducing it from the case $a = 1$, via a result about limits of constant multiples of functions), there would be a bit more scope for manipulating absolute values using the Triangle Inequality:

If $|x - a| \leqslant |a|/2$, then $$ |a| - |x| \leqslant ||a| - |x|| \leqslant |a - x| = |x - a| \leqslant |a|/2, $$ therefore $|x| \geqslant |a|/2$, therefore $$ \left\lvert\frac{1}{x} - \frac{1}{a}\right\rvert = \left\lvert\frac{a - x}{xa}\right\rvert = \frac{|a - x|}{|xa|} = \frac{|x - a|}{|x|\cdot|a|} \leqslant 2\frac{|x - a|}{|a|^2}. $$ Therefore, for all $\epsilon > 0$, $$ \text{if } 0 < |x - a| < \frac{|a|\cdot\min\left\{1,\, |a|\cdot\epsilon\right\}}{2}, \text{ then } \left\lvert\frac{1}{x} - \frac{1}{a}\right\rvert < \epsilon, $$ which isn't much more complicated than the special case $a = 1$.

I see that Michael Spivak, Calculus, 3rd ed. (1994), p.101, does it this way - so at least it's not wrong!

Answer 2

You have to solve $$ \left|\frac{1}{x}-1\right|<\varepsilon $$ and it's not restrictive to assume $x>0$ (that is, $\delta<1$ at the end). With this assumption, the inequality becomes $$ -\varepsilon x<1-x<\varepsilon x $$ so $$ \begin{cases} 1<x(1+\varepsilon) \\[6px] x(1-\varepsilon)<1 \end{cases} $$ Since it's not restrictive to assume $\varepsilon<1$, this becomes $$ \frac{1}{1+\varepsilon}<x<\frac{1}{1-\varepsilon} $$ which is indeed a neighborhood of $1$. If you really need a $\delta$, you can observe that $$ \frac{1}{1+\varepsilon}=1-\frac{\varepsilon}{1+\varepsilon}, \qquad \frac{1}{1-\varepsilon}=1+\frac{\varepsilon}{1-\varepsilon} $$ so you can take $$ \delta=\min\left\{\frac{1}{2},\frac{\varepsilon}{1+\varepsilon}, \frac{\varepsilon}{1-\varepsilon}\right\} $$ (where $1/2$ takes care of $\delta<1$).

Answer 3

If $ \lim_{x\to A}f(x)=L\ne 0 $ then $ \lim_{x\to A}1/f(x)=1/L. $

Proof: Let $ d>0 $ such that $ |x-A|<d\implies |f(x)-L|<\frac {|L|}{2} \implies |f(x)|>\frac {|L|}{2}\implies |\frac {1}{f(x)L}|<\frac {2}{|L|^2}. $

For any $ e>0 $ there exists $ d'\in (0,d) $ such that $ |x-A|<d'\implies |f(x)-L|<\frac {eL^2}{2}. $

Therefore for any $e>0$ there exists $d'\in (0,d)$ such that $$|x-A|<d' \implies |\frac {1}{f(x)}-\frac {1}{L}| =\frac {|f(x)-L|}{|f(x)L|}\leq |f(x)-L|\cdot \frac { 2}{L^2}<\frac {eL^2}{2}\cdot \frac {2}{L^2}=e. $$

In particular, with $f(x)=x$ and $A=1=L$ we have $\lim_{x\to 1}x=1$, so $\lim_{x\to 1}1/x=1.$

Remarks: (1). The idea is that when $x$ is close enough to $A ,$ the numerator in $\frac {|f(x)-L|}{|f(x)L|}$ can be as small as desired, while the denominator is "bounded away from $0$".....(2). In some cases we only wish to consider $x\to A$ with $x\ne A,$ in which case we change $|x-A|<d$ and $|x-A|<d'$ in the proof to, respectively, $0<|x-A|<d$ and $0<|x-A|<d'.$