Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
lim_{t
ightarrow0}left(\frac{4}{t}-\frac{4}{t^{2}+t}
ight)

Explanation:

Step1: Find a common denominator

First, find a common - denominator for the two fractions. The common denominator of $t$ and $t^{2}+t=t(t + 1)$ is $t(t + 1)$.
$\frac{4}{t}-\frac{4}{t^{2}+t}=\frac{4(t + 1)}{t(t + 1)}-\frac{4}{t(t + 1)}$

Step2: Combine the fractions

Combine the two fractions:
$\frac{4(t + 1)-4}{t(t + 1)}=\frac{4t+4 - 4}{t(t + 1)}=\frac{4t}{t(t + 1)}$

Step3: Simplify the fraction

Cancel out the common factor $t$ (since $t
eq0$ when taking the limit as $t
ightarrow0$):
$\frac{4t}{t(t + 1)}=\frac{4}{t + 1}$

Step4: Evaluate the limit

Now, find $\lim_{t
ightarrow0}\frac{4}{t + 1}$. Substitute $t = 0$ into $\frac{4}{t + 1}$.
$\lim_{t
ightarrow0}\frac{4}{t + 1}=\frac{4}{0 + 1}=4$

Answer:

$4$