Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
\\(\lim_{t\to0}(\frac{3}{t}-\frac{3}{t^{2}+t})\\)

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 $t(t + 1)$. So, $\frac{3}{t}-\frac{3}{t^{2}+t}=\frac{3(t + 1)-3}{t(t + 1)}$.

Step2: Simplify the numerator

Expand and simplify the numerator: $3(t + 1)-3=3t+3 - 3=3t$. So the expression becomes $\frac{3t}{t(t + 1)}$.

Step3: Cancel out the common factor

Cancel out the common factor $t$ (since $t
eq0$ when taking the limit as $t
ightarrow0$). We get $\frac{3}{t + 1}$.

Step4: Evaluate the limit

Now, find $\lim_{t
ightarrow0}\frac{3}{t + 1}$. Substitute $t = 0$ into $\frac{3}{t+1}$, we have $\frac{3}{0 + 1}=3$.

Answer:

$3$