Question

b) (limlimits_{t \to 0^+} left( \frac{1}{t} - \frac{3}{2t^2 + 3t}
ight))

Explanation:

Step1: Simplify the expression

First, find a common denominator for the two fractions. The common denominator of \( t \) and \( 2t^2 + 3t \) is \( t(2t + 3) \) (since \( 2t^2 + 3t = t(2t + 3) \)).
Rewrite the fractions with the common denominator:
\[
\frac{1}{t} - \frac{3}{2t^2 + 3t} = \frac{2t + 3}{t(2t + 3)} - \frac{3}{t(2t + 3)}
\]

Step2: Combine the fractions

Subtract the numerators:
\[
\frac{2t + 3 - 3}{t(2t + 3)} = \frac{2t}{t(2t + 3)}
\]

Step3: Cancel out common factors

Cancel out the common factor \( t \) (since \( t \to 0^+ \), \( t
eq 0 \) so we can cancel):
\[
\frac{2t}{t(2t + 3)} = \frac{2}{2t + 3}
\]

Step4: Evaluate the limit

Now, take the limit as \( t \to 0^+ \):
\[
\lim_{t \to 0^+} \frac{2}{2t + 3}
\]
Substitute \( t = 0 \) into the expression:
\[
\frac{2}{2(0) + 3} = \frac{2}{3}
\]

Answer:

\(\frac{2}{3}\)