Which function grows faster
$𝑓(𝑛)= 2^{𝑛^2+3𝑛}$ and $𝑔(𝑛) = 2^{𝑛+1}$
by using the limit theorem I will first simplify
then I will just get $$\lim_{n \to \infty} \dfrac{2^{n^2+3n}}{2^{n+1}}=\lim_{n \to \infty} 2^{n^2+3n-n-1}=\lim_{n \to \infty} 2^{n^2+2n-1}=\infty$$
Is this enough?
I say it will go then to infinity so the $f(n)$ is growing faster? I am asking this question because I have to find it by using limit but I didn't need to use l'hopital rule!
Best Answer
Before Edit: Your idea was correct, but you didn’t simplify the limit properly. $$\lim_{n \to \infty} \frac{2^{n^2+3n}}{2^n+1}$$ It is enough to divide both the numerator and denominator by $2^n$. $$\lim_{n \to \infty} \frac{\frac{2^{n^2+3n}}{2^n}}{\frac{2^n+1}{2^n}} = \lim_{n \to \infty} \frac{2^{n^2+3n-n}}{2^{n-n}+\frac{1}{2^n}} = \lim_{n \to \infty} \frac{2^{n^2+2n}}{1+\frac{1}{2^n}}$$ As $n \to \infty$, it becomes clear that the limit tends to $\infty$ since the numerator tends to $\infty$ while the denominator tends to $1$.
After Edit: Yes, your way is correct.