Divide numerator and denominator by the highest power of x in the denominator to find the limit.
-$$\lim _ { t \rightarrow \infty } \frac { \sqrt { 9 t ^ { 2 } - 27 } } { t - 3 }$$
A) does not exist
B) 3
C) 9
D) 27

Question

Quizplus · Accepted Answer

The highest power of $t$ in the denominator is 1, so we divide both the numerator and denominator by $t$. This gives us:
$$\lim_{t 	o \infty} \frac{\sqrt{9t^2 - 27}}{t-3} = \lim_{t 	o \infty} \frac{\sqrt{9 - \frac{27}{t^2}}}{1-\frac{3}{t}}$$
As $t$ approaches infinity, the fraction $\frac{27}{t^2}$ goes to zero, and the fraction $\frac{3}{t}$ goes to zero. So the limit simplifies to:
$$\lim_{t 	o \infty} \frac{\sqrt{9}}{1} = 3$$
Therefore, the answer is B.

Divide Numerator and Denominator by the Highest Power of X lim⁡t→∞9t2−27t−3\lim _ { t \rightarrow \infty } \frac { \sqrt { 9 t ^ { 2 } - 27 } } { t - 3 }t→∞lim​t−39t2−27​​

Divide Numerator and Denominator by the Highest Power of X $\lim _ { t \rightarrow \infty } \frac { \sqrt { 9 t ^ { 2 } - 27 } } { t - 3 }$