evaluate the limit\n\\(\\lim_{t\\to1}\\frac{\\frac{1}{t}-1}{\\sqrt{t}-1}\\)

Explanation:

Step1: Combine fractions in the numerator

First, find a common - denominator for the fractions in the numerator. The expression is $\lim_{t\rightarrow1}\frac{\frac{1}{t}-1}{1 - \sqrt{t}}=\lim_{t\rightarrow1}\frac{\frac{1 - t}{t}}{1-\sqrt{t}}=\lim_{t\rightarrow1}\frac{1 - t}{t(1 - \sqrt{t})}$.

Step2: Factor the difference - of - squares

Since $1 - t=(1-\sqrt{t})(1 + \sqrt{t})$, the limit becomes $\lim_{t\rightarrow1}\frac{(1-\sqrt{t})(1 + \sqrt{t})}{t(1 - \sqrt{t})}$.

Step3: Cancel out common factors

Cancel out the common factor $(1-\sqrt{t})$ (for $t\neq1$). We get $\lim_{t\rightarrow1}\frac{1+\sqrt{t}}{t}$.

Step4: Substitute $t = 1$

Substitute $t = 1$ into the simplified function: $\frac{1+\sqrt{1}}{1}=\frac{1 + 1}{1}=2$.

Answer:

$2$