Wednesday, February 27, 2013

Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 26

Determine the $\displaystyle \lim \limits_{t \to 0} \left( \frac{1}{t} - \frac{1}{t^2 + t} \right)$, if it exists.


$
\begin{equation}
\begin{aligned}
\lim \limits_{t \to 0} \left( \frac{1}{t} - \frac{1}{t^2 + t} \right)
& = \lim \limits_{t \to 0} \frac{t^2 + \cancel{t} - \cancel{t}}{t(t^2 + t)} = \lim \limits_{t \to 0} \frac{t^2 }{t(t^2 + t)}
&& \text{ Get the LCD and combine like terms}\\
\\
& = \lim \limits_{t \to 0} \frac{\cancel{t^2}}{\cancel{t^2} (t + 1) } = \lim \limits_{t \to 0} \frac{1}{t + 1}
&& \text{ Factor the denominator and cancel out like terms}\\
\\
& = \frac{1}{0 + 1} = \frac{1}{1}
&& \text{ Substitute value of $t$ and simplify}\\
\\
& = 1


\end{aligned}
\end{equation}
$

No comments:

Post a Comment