# When using the method of substituting variables to find the limit, why can we still use the original numbers that need to be approximated?

I'm trying to solve this problem:

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

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

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!

• The trick is to realize that the function is $f(t) = t^2 - 2$ on the corresponding domain. The answer is very horrendously phrased. Commented May 22 at 5:13
• @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! Commented May 22 at 5:32
• Correct. $\lim_{x\rightarrow 3} f(x) = \lim_{y \rightarrow 3 } f(y)$. $\quad$ What the solution is trying to express is: $\lim_{x + \frac{1}{x} \rightarrow 3 } f(x + \frac{1}{x} ) = \lim_{x + \frac{1}{x} \rightarrow 3} x^2 + \frac{1}{x^2} = \lim_{x + \frac{1}{x} \rightarrow 3} (x + \frac{1}{x})^2 - 2 = 7$. Commented May 22 at 5:35
• @CalvinLin I understand the problem, thank you for your help! :D Commented May 22 at 5:42
• @gidds Mainly familiarity with algebraic manipulation. EG It is known that $x^n + 1/x^n$ can be written as a polynomial in $x + 1/x$. Commented May 22 at 18:29

What the solution is trying to express is:

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

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. :)

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