How to take limits of inverse functions!

**kwonypoo** · June 17th 2012, 01:36 PM

]How would you solve

lim as x -> infinity f^-1( x² + 2 x + 1)

?

Thanks!

**pickslides** · June 17th 2012, 02:06 PM

Why not just find $f^{-1} (x^2+2x+1)$ ?

**skeeter** · June 17th 2012, 02:08 PM

$f(x) = x^2 +2x + 1 = (x+1)^2$

restricting the domain of f(x) to x > -1

$f^{-1}(x) = \sqrt{x} - 1$

the inverse is unbounded as $x \to \infty$

I'm hoping you don't have confusion between inverse and reciprocal functions.

**HallsofIvy** · June 17th 2012, 05:11 PM

What you wrote was $\lim_{x\to\infty} f^{-1}(x^2+ 2x+ 1)$ which we cannot help you with because you did not tell us what "f" is. Skeeter assumed you meant that $f(x)= x^2+ 2x+ 1$ and that you want to find $\lim_{x\to\infty} f^{-1}(x)$ . Is that correct?

**skeeter** · June 17th 2012, 05:14 PM

Originally Posted by HallsofIvy

What you wrote was $\lim_{x\to\infty} f^{-1}(x^2+ 2x+ 1)$ which we cannot help you with because you did not tell us what "f" is. Skeeter assumed you meant that $f(x)= x^2+ 2x+ 1$ and that you want to find $\lim_{x\to\infty} f^{-1}(x)$ . Is that correct?

that is another reason for my final statement ...

**Prove It** · June 17th 2012, 05:15 PM

I'm interested to know why this Algebra/Pre Calculus question was posted in Statistics...

**skeeter** · June 17th 2012, 05:33 PM

Originally Posted by Prove It

I'm interested to know why this Algebra/Pre Calculus question was posted in Statistics...

not like it matters anymore ... the new owner/moderator hasn't been around for awhile. just note how many problems have been posted in the lobby.

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