Suppose that f is a function such that lim[x->infty][f(x)+2x^2] = 1. Determine...

[kylej]

Suppose that f is a function such that

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \left[f(x) + 2x^2\right]\, = 1\)

Determine, if possible,

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \dfrac{f(x)}{x^2\, +\, 1}\)

Justify your answer.

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Steps needed.

 

[JeffM]

Suppose that f is a function such that

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \left[f(x) + 2x^2\right]\, = 1\)

Determine, if possible,

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \dfrac{f(x)}{x^2\, +\, 1}\)

Justify your answer.

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Steps needed.

Let's start by determining what f(x) might be. We need to get rid of that \(\displaystyle 2x^2.\)

The obvious possibility is

\(\displaystyle \displaystyle f(x) = 1 - 2x^2 \implies f(x) + 2x^2 = 1 \implies \lim_{x \rightarrow \infty}(f(x) + 2x^2) = \lim_{x \rightarrow \infty}(1) = 1.\)

Now what?

 

[HallsofIvy]

Is your question "how can we look at this and instantly know the answer without doing any calculation at all?" Gosh I can't answer that! I would have to actually compute things.

For example, it occurs to me that if we take \(\displaystyle f(x)= 1- 2x^2\) then \(\displaystyle \lim_{x\to\infty} f(x)+ 2x^2= 1\). Then \(\displaystyle \frac{f(x)}{x^2+ 1}= \frac{1- 2x^2}{x^2+ 1}\). Divide both numerator and denominator by \(\displaystyle x^2\) to get \(\displaystyle \frac{\frac{1}{x^2}- 2}{1+ \frac{1}{x^2}}\). Now what happens as x goes to infinity?