I am quite a newbie in calculus and I would like to know if this attempt to prove variable substitution in limits is rigorous enough. Since my goal is to learn from my mistakes I would appreciate any suggestions/improvements in order to get better at this.


$\lim_{x\to a} g(x)=b\quad(1)$

$\lim_{y\to b} f(y)=l\quad(2)$

Exists a deleted neighbourhood of $a$ where $g(x) \neq b\quad(3)$

$Im\ g \subseteq dom\ f$: it ensures we can safely write $f(g(x))$

$\Rightarrow \lim_{x\to a} f(g(x)) = l$ (so we successfully changed $y = g(x)$)


  • $\forall\epsilon_1\gt0,\ \exists\delta_1(\epsilon_1,a)\gt0:0\lt\lvert x-a\rvert\lt\delta_1\Rightarrow\lvert g(x)-b\rvert\lt\epsilon_1\quad(1)$
  • $\forall\epsilon_2\gt0,\ \exists\delta_2(\epsilon_2,b)\gt0: 0\lt\lvert y-b\rvert\lt\delta_2\Rightarrow\lvert f(y)-l\rvert\lt\epsilon_2\quad(2)$
  • $\exists\eta(a,b)>0: 0\lt\lvert x-a\rvert\lt\eta\Rightarrow g(x)\neq b\Leftrightarrow 0\lt\lvert g(x)-b\rvert\quad(3)$

Let $\epsilon_2\gt0$, so $\exists\delta_2\gt 0\Rightarrow 0\lt\lvert y-b\rvert\lt\delta_2\Rightarrow\lvert f(y)-l\rvert\lt\epsilon_2$ and choose $\epsilon_1 = \delta_2$ (since we know it is greater than zero). So $\exists\delta_1: 0\lt\lvert x-a\rvert\lt\delta_1\Rightarrow\lvert g(x)-b\rvert\lt\epsilon_1$. Pick $\delta_1' = \min\{\delta_1,\eta\}$ therefore $0\lt\lvert x-a\rvert\lt\delta_1'\Rightarrow 0\stackrel{\text{(3)}}{\lt}\lvert g(x)-b\rvert\stackrel{\text{(1)}}{\lt}\epsilon_1 = \delta_2$. We know from $(2)$ that $0\lt\lvert g(x)-b\rvert\lt\delta_2\Rightarrow \lvert f(g(x))-l\rvert\lt\epsilon_2$.

Thus, $\forall\epsilon_2\gt0\ \exists\delta_1'(\epsilon_2,\eta,a)\gt0: 0\lt\lvert x-a\rvert\lt\delta_1'\Rightarrow \lvert f(g(x))-l\rvert\lt\epsilon_2\quad\square$


Well done! The proof is very well written and presented.

It's also clear from the proof where you used all the properties, and you can also see from the proof that the third property is vital for the proof as the proof could not be completed without it (and in fact, taking a discontinuous $f$ and a constant $g$ can lead to a counterexample)

  • $\begingroup$ Thanks! If someone else reads this and is wondering what the counterexample is, I'll leave this link here (which also provides a different proof): limitinfinite.org/… $\endgroup$
    – J. Doe
    Jun 12 '17 at 13:41

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