This is no homework. It's another task of a sample exam and I'd like to know how to solve it.

Find the limit of $$\lim_{n\to \infty}\frac{1+(\sqrt{n}+1)^{3}+2\sqrt{n}}{n+\sin(n)}$$

Both numerator and denominator go towards $\infty$ for $n \rightarrow \infty$. So I have tried using L'Hôpitals rule because we got $\frac{\infty}{\infty}$.

Here I had the first question in mind, shall I simplify everything before I derivate? (In most cases I think that would be very useful...?)

I have tried both ways (simplify before derivate, derivate and then simplify) and both ended up in $\frac{\infty}{\infty}$ even after several derivations.

Here is how I simplified it:

$$\lim_{n\rightarrow \infty}\frac{n\sqrt{n}+3n+5\sqrt{n}+2}{n+\sin(n)}$$

Then I differentiated the numerator and denominator:

$f'(x) = \frac{3}{2}\sqrt{n}+3+\frac{5}{2\sqrt{n}}$, $g'(x) = 1+\cos(n)$


$$\lim_{n\rightarrow \infty}\frac{\frac{3}{2}\sqrt{n}+3+\frac{5}{2\sqrt{n}}} {1+\cos(n)}$$

Doesn't really help me...

  • 1
    $\begingroup$ Note that you can also see your L'Hopitaled limit diverge off to infinity since the denominator is bounded by $2$ whilst the numerator goes off to infinity. $\endgroup$
    – Zain Patel
    Jul 23, 2016 at 12:17

1 Answer 1


Expand the cube and divide top and bottom by $n$ to get $$\begin{align*}\lim_{n \to \infty} \frac{n\sqrt{n} +3n + 5\sqrt{n} + 2}{n + \sin n} &= \lim_{n \to \infty} \frac{\sqrt{n} + 3 + \frac{5}{\sqrt{n}} + \frac{2}{n}}{1 + \frac{\sin n}{n}} \end{align*}$$ from which you can see the numerator diverges off to $\infty$ whilst the denominator tends to $1$. So the limit diverges off to infinity as whole.

To see that the denominator tends to $1$, it suffices to note that $\frac{\sin n}{n} \to 0$ this is demonstrated via $-1 \leq \sin n \leq 1 \Rightarrow -\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}$ and then squeezing.

Intuitively, this is to be expected since the highest power of $n$ in the numerator $n^{3/2}$ grows faster than the highest power of $n$ in the denominator $n^1$ whilst $|\sin n|$ is bounded by $1$. So the numerator grows too wildly for the limit to be finite.

  • $\begingroup$ it'actually a cube but your argument and the end result remain correct. $\endgroup$
    – Frank
    Jul 23, 2016 at 12:12
  • $\begingroup$ @FrancescoAlem. thank you, fixed. I need new glasses(!). $\endgroup$
    – Zain Patel
    Jul 23, 2016 at 12:14
  • $\begingroup$ Oh well I was on the right track at least, a bit :P Thank you very much for your quick answer. One question, you said denominator tends to 1. I cannot just say that I think. So it would be allowed to pick it and count this limit (the denominator only) and then continue? $\endgroup$
    – cnmesr
    Jul 23, 2016 at 12:18
  • 1
    $\begingroup$ Sure, the denominator is $\lim_{n\to \infty}1 + \frac{\sin n}{n}$. This is $\lim_{n \to \infty} 1 + \lim_{n \to \infty} \frac{\sin n}{n} = 1 + \lim_{n\to \infty} \frac{\sin n}{n}$. Now it is basic that $|\sin n|$ is bounded above by $1$ whilst $n$ goes off to infinity so $-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}$ and taking the limit (squeezing) gives $\frac{\sin n}{n} \to 0$ so the entire denominator is just $1 + 0 = 1$. $\endgroup$
    – Zain Patel
    Jul 23, 2016 at 12:20

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