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Why does $\lim_{x \rightarrow a} \cos (x-a)=\lim_{x \rightarrow 0} \cos (x)$ without using continuity of $\cos$ function?

In general when is it okay to "switch" the limit like this. There is obviously something going on that I am not aware about.

Could anyone explain.

Thanks.

Sam
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    Just note that while $x\to a$ indeed $(x-a)\to 0$ – Mikasa Dec 09 '15 at 11:16
  • Okay that makes sense but is that rigorous enough to just write that and then switch it? – Sam Dec 09 '15 at 11:21
  • "Rigorous enough" is context-dependent. If you're in physics or calculus, probably just making the claim is fine. If you're studying analysis and have just learned the formal definition of limits, it's likely you're expected to verify using the definition. – Andrew D. Hwang Dec 09 '15 at 11:26
  • Yeah I'm in analysis. But I don't exactly understand what is going on when it changes from the definition side of things. Could you explain please? – Sam Dec 09 '15 at 11:31
  • @Sam: No time at the moment, but to start, write down the formal definitions (using $\varepsilon$'s and $\delta$'s) of "$\lim\limits_{x \to a} \cos(x - a) = L$" and "$\lim\limits_{u \to 0} \cos(u) = L$" as separate logical conditions, and use the "substitution" $u = x - a$ to show each condition implies the other. – Andrew D. Hwang Dec 09 '15 at 11:48
  • Okay thanks I see what you mean using H.R's answer. I got it now. – Sam Dec 09 '15 at 11:52

1 Answers1

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Definition of limit of a function $f(x)$ at point $0$ is

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x} \right| < \delta \to \left| {f(x) - L} \right| < \varepsilon } \right)} \right)$$

Now, if we rewrite the definition

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {(x+a)-a} \right| < \delta \to \left| {f((x+a)-a) - L} \right| < \varepsilon } \right)} \right)$$

and set $y=x+a$, we will get

$$\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall y,0 < \left| {y-a} \right| < \delta \to \left| {f(y-a) - L} \right| < \varepsilon } \right)} \right)$$

and hence

$$\lim_{x \to 0} f(x) =\lim_{y \to a} f(y-a)$$

Hosein Rahnama
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  • Great answer. I understand this case now, you just change the variable and it follows from the definition. Is there a more general argument for when you can switch the variable in other cases I wonder? – Sam Dec 09 '15 at 11:53
  • @Sam: The most general case is the composition theorem for limits, take a look at that. :) – Hosein Rahnama Dec 09 '15 at 11:55
  • Will do thanks. – Sam Dec 09 '15 at 11:55