Question

evaluate the limit: $lim_{t
ightarrow0}left(\frac{5}{tsqrt{25 + t}}-\frac{1}{t}
ight)=$

Explanation:

Step1: Combine the fractions

First, find a common - denominator. The common denominator of \(t\sqrt{25 + t}\) and \(t\) is \(t\sqrt{25 + t}\). So \(\frac{5}{t\sqrt{25 + t}}-\frac{1}{t}=\frac{5-\sqrt{25 + t}}{t\sqrt{25 + t}}\).

Step2: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator \(5+\sqrt{25 + t}\).
\[

$$\begin{align*} &\frac{(5 - \sqrt{25 + t})(5+\sqrt{25 + t})}{t\sqrt{25 + t}(5+\sqrt{25 + t})}\\ =&\frac{25-(25 + t)}{t\sqrt{25 + t}(5+\sqrt{25 + t})}\\ =&\frac{25 - 25 - t}{t\sqrt{25 + t}(5+\sqrt{25 + t})}\\ =&\frac{-t}{t\sqrt{25 + t}(5+\sqrt{25 + t})} \end{align*}$$

\]

Step3: Simplify the expression

Cancel out the common factor \(t\) in the numerator and denominator. We get \(\frac{-1}{\sqrt{25 + t}(5+\sqrt{25 + t})}\).

Step4: Evaluate the limit

Now, find \(\lim_{t
ightarrow0}\frac{-1}{\sqrt{25 + t}(5+\sqrt{25 + t})}\). Substitute \(t = 0\) into the expression: \(\frac{-1}{\sqrt{25+0}(5+\sqrt{25 + 0})}=\frac{-1}{5\times(5 + 5)}=-\frac{1}{50}\).

Answer:

\(-\frac{1}{50}\)