Question

If $f\left(x\right)=\frac{3x+12}{2x-12}$, find the value to which $f\left(x\right)$ approaches as $x$ tends to positive infinity.

• A. $-6$
• B. $-\frac{3}{2}$
• C. $-1$
• D. $\frac{2}{3}$
• E. $\frac{3}{2}$

Answer 1

The correct option is E $\frac{3}{2}$

Given, $f\left(x\right)=\frac{3x+12}{2x-12}$

We can write the above equation into

$f\left(x\right)=\frac{\frac{1}{x}(3x+12)}{\frac{1}{x}(2x-12)}=\frac{3+\frac{12}{x}}{2-\frac{12}{x}}$

$x$ gets infinitely larger, the expression $\frac{12}{x}$ approches the value $0$, so as $x$ gets infinetely larger, the expression $\frac{3+\frac{12}{x}}{2-\frac{12}{x}}$ approches the value $\frac{3+0}{2+0}=\frac{3}{2}$

Therefore, as $x$ gets infinitely larger, $f\left(x\right)$ approches $\frac{3}{2}$.