Prove that $f: [0, \infty) \to \mathbb{R}$ given by $f(x) = \frac{1+2^x+x^3}{4+5e^x}$ is bounded.

Question

Prove that $f: [0, \infty) \to \mathbb{R}$ given by $f(x) = \frac{1+2^x+x^3}{4+5e^x}$ is bounded.

Is it enough to just prove that the limit is $0$ as x approaches infinity? I'm thinking not but don't know how to approach the problem.

Answer 1

Assuming you meant $f:[0,\infty)\to\mathbb{R}$, proceed as follows. By definition, if $\lim_{x\to\infty}f (x)=0$, then for all $\varepsilon>0$ there exists some $N>0$ such that if $x>N$, then $|f(x)|<\varepsilon$. So if you fix $\varepsilon$ to be some value, say $1$, then you have that there exists an $N_0$ such that $|f(x)|<1$ on $(N_0,\infty)$.

Now if you also prove that $f$ is continuous on $[0,\infty)$, then it must also be continuous on $[0,N_0]$. But this is a compact set, and so $f$ is bounded on $[0,N_0]$. Make sure you are able to justify why this is true.

Combining these, you've shown that $f$ is bounded on $[0,N_0]$ and $(N_0,\infty)$. Conclude that $f$ is bounded on $[0,\infty)$.

Answer 2

Explicit bound for $f$: $0\leq f(x) \leq \frac { 1+2^{x}+x^{3}} {5e^{x}}$. Note that $\frac 1 {5e^{x}}<\frac 1 5$, $\frac {2^{x}} {5e^{x}}<\frac 1 5$ (because $\frac 2 5 <1$) and $\frac {x^{3}} {5e^{x}} \leq \frac {x^{3}} {5x^{3}/(3!)}=\frac 6 5$. Hence $0 \leq f(x) \leq \frac 8 5$ for all $x$.