0
$\begingroup$

I'm learning Silverman's "Modern calculus and analytic geometry", on page 166 example 2, there is a proof of $$\lim_{x \to 1}\frac{1}{x^2-1}=\infty$$ The original text is:

The function $$f(x)=\frac{1}{x^2-1} \qquad(x\neq \pm 1)$$ approaches $\infty$ as $x\to 1$. In fact, given any M>0, choose $\delta = \min \{1/3M, 1\}$. Then $0<|x-1|<\delta$ implies $1<|x+1|<3$ and (...)

(If needed, I will copy the whole example here.)

My silly first question is does the "1/3M" in "$\min\{1/3M, 1\}$" means $\frac {M}{3}$ or $\frac{1}{3M}$ ?

And since I won't expect to figure out where the set {1/3M, 1} come from in the next hours, I will be very glad if you tell me.

  • 1
    Euh, the left limit is $-\infty$ whereas the right limit is $+\infty$. The limit doesn't exist unless you further specify the domain. – 2017-01-24
  • 1
    They mean $\frac1{3M}$. $\delta$ is supposed to get smaller the larger $M$ is, so that's the only thing that makes sense. – 2017-01-24

1 Answers 1

1

I believe it means $\frac{1}{3M}$.

You're trying to show that for every $M>0$, there is a $\delta>0$ such that if $|x-1|<\delta$, then $\frac{1}{x^2-1}>M$. Notice that by the Triangle Inequality, $x<1+\delta$. Substituting and expanding we get $$|\frac{1}{x^2-1}|=\frac{1}{|x-1||x+1|}>\frac{1}{\delta|x+1|}>\frac{1}{\delta|(1+\delta)+1|}=\frac{1}{\delta(2+\delta)}>\frac{1}{3\delta}=\frac{1}{3\,(1/(3M))}=M.$$ Therefore $\lim_{x\to1}|\frac{1}{x^2-1}|=+\infty$. As @Arthur pointed out this only works for the absolute value of the function, so this is only a partial answer (though it does prove it is unbounded above).

  • 0
    This is a good answer, but it still doesn't address the main problem with the question formulation, namely that $x^2-1$ could be negative. For instance, you've just assumed $x^2-1=|x+1||x-1|$ when that's not true for $x<1$. Note, if we approach $1$ _from above_, your solution is exemplary. – 2017-01-24