By using the definition of limit ONLY, prove that $$ \lim_{x\to 1^+} \dfrac{x}{x-1} = +\infty $$ (Note that you need to use the definition of one-sided limit from the right and you are NOT allowed to use any algebra or any theorem for limit.)

I keep getting mixed up with the definitions when trying to answer this, when attempting the question i've started with

For all $M\in\mathbb{R}$ we need to find a $\delta>0$ such that $\dfrac{1}{1-x}>M$ for all $x\in\mathbb{R}$ and $0<x<\delta$

Not too sure what to do from here or if this is even correct.

  • 1
    $\begingroup$ The way you interpret the limit by definition is terribly wrong. Checking it in your book again would help. $\endgroup$ – user9464 Aug 6 '17 at 21:30
  • $\begingroup$ Where did the $\frac 1{1-x}$ come from? And if the condition so that $0 <x < \delta$ how are you taking that $x \rightarrow 1^+$ into account? If you have to find it for $0<x < \delta$ having nothing to do with $1^+$ that would mean all limits are equal because in makes no difference what $x \rightarrow$ towards... So, recheck the definition of limits (or of infinite limits). $\endgroup$ – fleablood Aug 6 '17 at 22:08
  • $\begingroup$ This question only makes sense if you extend the real line by adjoining an 'infinity' element (i.e. one point compactification). On the real line, the limit definition only applies to finite limiting values. $\endgroup$ – Aurel Aug 7 '17 at 14:43


$$\frac {x}{x-1}=1+\frac {1}{x-1} $$

We need prove that

$$(\forall A>2 )\;\;\;(\exists\eta>0)\;\;(\forall x>1)$$ $$0 <x-1 <\eta \implies 1+\frac {1}{x-1}>A $$

this last condition is equivalent to

$$ x-1 <\frac {1}{A-1}.$$

So we can take $\eta=\frac {1}{A-1} $.


Set $y = x-1$, then:

$\lim_{y \rightarrow 0^+} \frac{y+1}{y} = + \infty$, or

$\lim_{y \rightarrow 0^+} ( 1 + \frac{1}{y}) = + \infty$.

Let $M \in \mathbb{R^+}$ be given.

Choose $0 \lt \delta \lt 1/M$.

For $0 \lt y \lt \delta$:

$ M \lt 1 + \frac{1}{y}$.


The actual definition of $\lim_{x \rightarrow 1^+}\frac x{x-1} = +\infty$ is actually:

For any $M \in \mathbb R$ one can find a $\delta$ (probably dependent upon the value of $M$) so that whenever $0 < |x- 1| < \delta$ and $x > 1$ then $\frac x{x-1} > M$.

This is different than what you wrote.

Use algebra to solve if $x-1 < ????what???$ and $x > 1$ then $\frac x{x-1} > M$.

Hint: If $x-1< \delta$ then $\frac x{x-1} > \frac x {\delta} > \frac 1{\delta}$.

So when is $\frac 1{\delta} > M$?


Note that $$ \frac{x}{x-1} > \frac{1}{x-1} $$ if $0 < x-1 < 1$. Given any $M > 0$, we have $$ \frac{1}{x-1} > M $$ if in addition $0 < x-1 < 1/M$. So $0 < x-1 < \min \{ 1, 1/M \}$ implies $\frac{x}{x-1} > M$. Taking $\delta := \min \{ 1, 1/M \}$ suffices.


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