Limit problem with function with bounded derivative Let $f:\mathbb{R}\to \mathbb{R}$ be a differentiable function with bounded first derivative and $g(x)=x^2+f(x)$.
Find following limits:
$$ \lim_{x \to +\infty}g(x)$$ and
$$ \lim_{x\to +\infty}\frac{g(x)}{\sqrt{1+x^4}}$$
My work:
If $f$ has bounded first derivative that means that $|f'(x)|<M$ where $M \in \mathbb{R}$. Applying Lagrange mean value theorem I've got $|f(x)-f(y)|=|f'(\epsilon)|\cdot |x-y|< M|x-y|$. This means that function is Lipschitz. Does that implies that $f$ have limit at infinity? If so, is following conclusion correct:
$$\lim_{x \to +\infty}g(x)=+\infty$$ because $x^2$ tends to $+\infty$, and
$$ \lim_{x\to +\infty}\frac{g(x)}{\sqrt{1+x^4}}=\lim_{x\to +\infty}\frac{g(x)}{x^2\sqrt{1+\frac{1}{x^4}}}=\lim_{x\to +\infty}\frac{g(x)}{x^2}=\lim_{x\to +\infty}\frac{x^2+f(x)}{x^2}=1$$
 A: If, for $x>x_0$, you have that $|f'(x)|< M$, for some constant $M>0$, the following inequalities hold
$$
f(x_0)- Mx \leq f(x) \leq f(x_0)+Mx.
$$
this means that
$$
\underbrace{x^2+f(x_0)-Mx}_{\to +\infty} \leq g(x) \leq \underbrace{x^2+f(x_0)+Mx}_{\to +\infty}
$$
and, consequently, $\lim_{x \to +\infty} g(x)=+\infty$.
Now, we can just use L'Hôpital's rule:
$$
\lim_{x\to +\infty} \frac{g(x)}{\sqrt{1+x^4}}=\lim_{x\to +\infty}\dfrac{2x+g'(x)}{\frac{4x^3}{2\sqrt{1+x^4}}}=\lim_{x\to +\infty}\frac{\sqrt{1+x^4}}{x^2}=\lim_{x\to +\infty}\sqrt{\frac{1}{x^4}+1}=1
$$
A: There is a simple example to show the answer to your first question is no: try $f(x) = \sin x$. This function has a bounded derivative but no limit at $\infty$.
On the other hand, the Lipschitz condition $|f(x) - f(y)| \le M|x-y|$ tells you that $|f(x) - f(0)| \le M|x|$ for any $x$, and thus $$x^2 + f(x) \ge x^2 + f(0) - M|x| \to \infty.$$
Likewise,
$$0 \le \frac{|f(x)|}{\sqrt{x^4 + 1}}\le \frac{|f(0)| + M|x|}{\sqrt{x^4 + 1}} \to 0.$$
