Find limlimits_(n→∞)|((n^2+3i)(n-i))/(in^3-3n+4-i)|
Find $\ \lim\limits_{n\to\infty}\left|\dfrac{(n^2+3i)(n-i)}{in^3-3n+4-i}\right|$.
Explanation of Method
Expand numerator and denominator into real/imaginary form, take moduli, then divide through by the dominant power $n^6$ inside each square root so that only the leading-order ($n^6$) terms survive in the limit.
Step 1 · Expand the numerator
$$ (n^2+3i)(n-i)=n^3-n^2i+3ni-3i^2=(n^3+3)+i(3n-n^2) $$
Step 2 · Rewrite the denominator
$$ in^3-3n+4-i=(4-3n)+i(n^3-1) $$
Step 3 · Compute each modulus
$$ |{\rm num}|^2=(n^3+3)^2+(3n-n^2)^2=n^6+6n^3+9+n^4-6n^3+9n^2=n^6+n^4+9n^2+9 $$
$$ |{\rm den}|^2=(4-3n)^2+(n^3-1)^2=9n^2-24n+16+n^6-2n^3+1=n^6-2n^3+9n^2-24n+17 $$
Step 4 · Factor out the dominant $n^6$
$$ \left|\frac{\rm num}{\rm den}\right|=\sqrt{\frac{n^6+n^4+9n^2+9}{n^6-2n^3+9n^2-24n+17}}=\sqrt{\frac{1+\frac1{n^2}+\frac9{n^4}+\frac9{n^6}}{1-\frac2{n^3}+\frac9{n^4}-\frac{24}{n^5}+\frac{17}{n^6}}} $$
Step 5 · Take the limit
Every fractional term vanishes as $n\to\infty$, leaving:
$$ \boxed{\lim_{n\to\infty}\left|\frac{(n^2+3i)(n-i)}{in^3-3n+4-i}\right|=\sqrt{\frac11}=1} $$
Shortcut · leading-term estimate
Numerator $\sim n^2\cdot n=n^3$ in magnitude; denominator $\sim in^3$, also magnitude $n^3$. The ratio of magnitudes tends to $1$ directly — matching the more careful computation above.
Result
Both numerator and denominator are degree-$3$ polynomials in $n$ (as complex expressions), so their moduli grow at exactly the same $n^3$ rate — the limit of the ratio is the ratio of leading coefficients' magnitudes, both equal to $1$.