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I'm trying to solve this problem:

Assume $$f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2}$$ then $$\lim_{x\to 3} f(x) = ?$$

I saw the answer is

$$t = x+\frac{1}{x}$$ $$\lim_{x\to 3} f(x) = \lim_{t\to 3} t^2 - 2 = 7$$

I am a little confused, since we have set $t=x+\frac{1}{x}$, why $x\to 3$ doesn't become $t\to \frac{10}{3}$?

This question may be a bit low-level, but I am really not good at math. Thank you for your help!

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    $\begingroup$ The trick is to realize that the function is $f(t) = t^2 - 2$ on the corresponding domain. The answer is very horrendously phrased. $\endgroup$
    – Calvin Lin
    Commented May 22 at 5:13
  • $\begingroup$ @CalvinLin You mean, as long as the relationship is $f$, no matter what the variables are, it is consistent with $x\to 3$? What I don't quite understand is why the numbers to be approximated don't change correspondingly in the process from $x+\frac{1}{x}$ to $x$. Thanks for your help! $\endgroup$ Commented May 22 at 5:32
  • $\begingroup$ Correct. $ \lim_{x\rightarrow 3} f(x) = \lim_{y \rightarrow 3 } f(y) $. $\quad$ What the solution is trying to express is: $\lim_{x + \frac{1}{x} \rightarrow 3 } f(x + \frac{1}{x} ) = \lim_{x + \frac{1}{x} \rightarrow 3} x^2 + \frac{1}{x^2} = \lim_{x + \frac{1}{x} \rightarrow 3} (x + \frac{1}{x})^2 - 2 = 7$. $\endgroup$
    – Calvin Lin
    Commented May 22 at 5:35
  • $\begingroup$ @CalvinLin I understand the problem, thank you for your help! :D $\endgroup$ Commented May 22 at 5:42
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    $\begingroup$ @gidds Mainly familiarity with algebraic manipulation. EG It is known that $x^n + 1/x^n$ can be written as a polynomial in $ x + 1/x$. $\endgroup$
    – Calvin Lin
    Commented May 22 at 18:29

2 Answers 2

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What the solution is trying to express is:

$$\lim_{x\rightarrow 3 } f(x) = \lim_{x + \frac{1}{x} \rightarrow 3 } f(x + \tfrac{1}{x} ) = \lim_{x + \frac{1}{x} \rightarrow 3} x^2 + \frac{1}{x^2} = \lim_{x + \frac{1}{x} \rightarrow 3} (x + \tfrac{1}{x})^2 - 2 = 7.$$

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There is an accepted answer, but I will write here another way to look at what's going on. You are given $$f\left(x+\frac1x\right)=x^2+\frac1{x^2}$$ and asked $\lim_{x\to3}f(x)$. Let me just switch variables then things will be more clear. Say we are given $$f\left(t+\frac1t\right)=t^2+\frac1{t^2}$$

Now, you want to find $\lim_{x\to3}f(x)$, which basically reads: "As the input of $f$ goes to $3$, what does the output go to?". You cannot have $x\to 3$ and $x+\frac1x\to3$ at the same time. The source of the confusion for you is simply using $x$ for two different things. Writing it with $t$ makes sense.

Anyway, once you show $f(t)=t^2-2$, you get the limit as $3^2-2$ (input is $x$ and $t+\frac1t$ in our notation and formula).

Hope this helps. :)

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  • $\begingroup$ Yeah you're right. I mistakenly confused parameters with unknowns. Thank you :) $\endgroup$ Commented May 22 at 13:20

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