# Evaluate? lim (5/(t√(1+t))-5/t)‎ t→0‎

## The answer must be -2.5, but I'm stuck because we have 5 which is a big number...plz explain when you solve it!

Feb 13, 2018

${\lim}_{t \rightarrow 0} \left(\frac{5}{t \sqrt{1 + t}} - \frac{5}{t}\right)$
plugging in 0 directly gives $\infty - \infty$, so more simplification is needed.

combine the fractions:
${\lim}_{t \rightarrow 0} \left(\frac{5}{t \sqrt{1 + t}} - \frac{5 \sqrt{1 + t}}{t \sqrt{1 + t}}\right)$
${\lim}_{t \rightarrow 0} \left(\frac{5 - 5 \sqrt{1 + t}}{t \sqrt{1 + t}}\right)$

conjugate the numerator:
${\lim}_{t \rightarrow 0} \left(\frac{\left(5 - 5 \sqrt{1 + t}\right) \left(5 + 5 \sqrt{1 + t}\right)}{t \sqrt{1 + t} \left(5 + 5 \sqrt{1 + t}\right)}\right)$
${\lim}_{t \rightarrow 0} \left(\frac{25 - 25 \left(1 + t\right)}{t \sqrt{1 + t} \left(5 + 5 \sqrt{1 + t}\right)}\right)$
${\lim}_{t \rightarrow 0} \left(\frac{25 - 25 - 25 t}{t \sqrt{1 + t} \left(5 + 5 \sqrt{1 + t}\right)}\right)$
${\lim}_{t \rightarrow 0} \left(\frac{- 25 t}{t \sqrt{1 + t} \left(5 + 5 \sqrt{1 + t}\right)}\right)$

divide t from the numerator and denominator:
${\lim}_{t \rightarrow 0} \left(\frac{- 25}{\sqrt{1 + t} \left(5 + 5 \sqrt{1 + t}\right)}\right)$

now plug in $t = 0$
${\lim}_{t \rightarrow 0} \left(- \frac{25}{\sqrt{1 + 0} \left(5 + 5 \sqrt{1 + 0}\right)}\right)$
${\lim}_{t \rightarrow 0} \left(- \frac{25}{\sqrt{1} \left(5 + 5 \sqrt{1}\right)}\right)$
${\lim}_{t \rightarrow 0} \left(- \frac{25}{5 + 5}\right)$
${\lim}_{t \rightarrow 0} \left(- \frac{25}{10}\right)$
${\lim}_{t \rightarrow 0} \left(- \frac{5}{2}\right)$
$= - 2.5$