Question

15 mark for review
lim(x→∞) (√(9x⁴ + 1))/(4x²+3)
a 1/3
b 3/4
c 3/2
d 9/4
e infinite

Explanation:

Step1: Divide numerator and denominator by $x^2$

As $x\to\infty$, we have $\lim_{x\to\infty}\frac{\sqrt{9x^{4}+1}}{4x^{2}+3}=\lim_{x\to\infty}\frac{\sqrt{x^{4}(9 + \frac{1}{x^{4}})}}{x^{2}(4+\frac{3}{x^{2}})}=\lim_{x\to\infty}\frac{x^{2}\sqrt{9+\frac{1}{x^{4}}}}{x^{2}(4 + \frac{3}{x^{2}})}$

Step2: Simplify the expression

Cancel out $x^{2}$ terms: $\lim_{x\to\infty}\frac{\sqrt{9+\frac{1}{x^{4}}}}{4+\frac{3}{x^{2}}}$

Step3: Evaluate the limit

As $x\to\infty$, $\frac{1}{x^{4}}\to0$ and $\frac{1}{x^{2}}\to0$. So $\lim_{x\to\infty}\frac{\sqrt{9+\frac{1}{x^{4}}}}{4+\frac{3}{x^{2}}}=\frac{\sqrt{9 + 0}}{4+0}=\frac{3}{4}$

Answer:

B. $3/4$