Find the limit. $$\lim _ { t \rightarrow \infty } \frac { t ^ { 2 } + 3 } { t ^ { 3 } + t ^ { 2 } - 7 }$$
A) $\infty$
B) -3
C) 0
D) 3
E) 7

Question

Quizplus · Accepted Answer

To find the limit, we can divide the numerator and denominator by the highest power of $t$ in the denominator, which is $t^3$. This gives $$\lim_{t\to\infty}\frac{t^2+3}{t^3+t^2-7}=\lim_{t\to\infty}\frac{\frac{t^2}{t^3}+\frac{3}{t^3}}{1+\frac{1}{t}-\frac{7}{t^3}}.$$As $t$ approaches infinity, all the terms with a $t$ in the denominator approach zero, so we are left with $$\frac{0+0}{1+0-0}=\boxed{0}.$$

Find the Limit lim⁡t→∞t2+3t3+t2−7\lim _ { t \rightarrow \infty } \frac { t ^ { 2 } + 3 } { t ^ { 3 } + t ^ { 2 } - 7 }t→∞lim​t3+t2−7t2+3​

Find the Limit $\lim _ { t \rightarrow \infty } \frac { t ^ { 2 } + 3 } { t ^ { 3 } + t ^ { 2 } - 7 }$