If $|f'(x)-e^{2x}|\leq 3$ for all $x\in\mathbb{R}$. Then evaluate the limit $\lim_{x\to +\infty}\frac{f(x)}{e^{2x}}$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function such that $f'$ is continuous and satisfies $|f'(x)-e^{2x}|\leq 3$ for all $x\in\mathbb{R}$. Then evaluate the limit $$\lim_{x\to +\infty}\frac{f(x)}{e^{2x}}$$.

My Attempt:

From the given inequality we have $$|f'(x)e^{-2x}-1|\leq 3e^{-2x}$$ $$\lim_{x\to+\infty}|f'(x)e^{-2x}-1|\leq\lim_{x\to+\infty} 3e^{-2x}$$ $$\lim_{x\to+\infty}|f'(x)e^{-2x}-1|\leq 0$$ $$\lim_{x\to+\infty}|f'(x)e^{-2x}-1|= 0$$ $$\lim_{x\to+\infty}\left(f'(x)e^{-2x}-1\right)= 0$$ $$\lim_{x\to+\infty}f'(x)e^{-2x}=1$$ Now to evaluate $\lim_{x\to +\infty}\frac{f(x)}{e^{2x}}$ it is very tempting to use the L'Hopital Rule and then using $\lim_{x\to+\infty}f'(x)e^{-2x}=1$ we get the answer as $\frac{1}{2}$

But how do we show that $\lim_{x\to+\infty}f(x)=\infty$ since condition to use L'Hopital rule is that the limit must be of form $\frac{0}{0}$ or $\frac{\infty}{\infty}$

Answer 1

You can avoid L'Hopital if you start with the inequality $f'(x) \le e^{2x} + 3$ and integrate on the interval $[0,x]$ to get $$f(x) - f(0) \le \frac{e^{2x}}{2} - \frac 12 + 3x.$$ In particular $$\frac{f(x)}{e^{2x}} \le \frac 12 + e^{-2x} \left( f(0) - \frac 12 + 3x \right)$$ for $x > 0$.

Now do something similar with the inequality $f'(x) \ge e^{2x} - 3$ and see what you can conclude.

Answer 2

If $f'(x) \geq e^{2x}-3$, then for $x \gt 0$, $$f(x) -f(0) = \int_0^x f'(t)~dt \geq \int_0^x (e^{2t}-3) ~dt=\frac{e^{2x}}{2}-3x-\frac 12.$$ Thus, $$f(x) \geq \frac{e^{2x}}{2}-3x+f(0)-\frac 12$$ and the right-hand side has limit $\infty$ as $x \to \infty$.

Answer 3

It's a common misunderstanding that L'Hopital requires both of $|f(x)|$ and $|g(x)|$ go to $\infty$ in order to evaluate $\lim \frac{f(x)}{g(x)}$. But $|f(x)|\rightarrow\infty$ is not used in the proof. This is stated explicitly in Wikipedia.