Prove that a given line is a horizontal asymptote for a given function

Question

So, here's the problem I'm trying to do:

Let $f(x) = \frac{x^2+2}{x^2+1}$. Prove that $f(x)$ has a horizontal asymptote given by $y = 1$.

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Proof Attempt:

This amounts to proving that either of the following two conditions hold:

$$\lim_{x \to +\infty} f(x) = 1$$

$$\lim_{x \to -\infty} f(x) = 1$$

So, we attempt proving the first limit. Let $\epsilon > 0$ be given. Then, we need an $M>0$ such that:

$$x > M \implies |f(x)-1| < \epsilon$$

$$|f(x)-1| = |\frac{1}{x^2+1}| \leq \frac{1}{x^2-1} < \epsilon$$

$$\frac{1}{\epsilon} < x^2 -1$$

$$\frac{1}{\epsilon} + 1 < x^2$$

$$x > \sqrt{\frac{1+\epsilon}{\epsilon}}$$

Define $M = \sqrt{\frac{1+\epsilon}{\epsilon}}$. Since we have our required $M$, we have shown that $\lim_{x \to +\infty} f(x) = 1$ and that the line $y = 1$ is a horizontal asymptote of $f(x)$

Does the argument above work? If it doesn't, why? How can I fix it?

Answer 1

$f(x)=\frac{x^2+2}{x^2+1}$ is even, $f(x)=f(-x)$, suffices to consider $\lim \rightarrow \infty$.

$x>1$;

$|f(x)-1|= |\frac{1}{x^2+1}|<\frac{1}{x^2} <\frac{1}{x}.$

Let $\epsilon >0$ be given;

Choose $M=1/(\epsilon)$.

For $x > M(>1)$ we have

$|f(x)-1| < \frac{1}{x} <1/M =\epsilon$.