Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 7

Determine the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $( 1, 1)$

$\text{Using the definition (Slope of the tangent line)}$

$\displaystyle m = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x -a}$

We have $a = 1$ and $f(x) = \sqrt{x}$, so the slope is


$
\begin{equation}
\begin{aligned}

\displaystyle m =& \lim \limits_{x \to 1} \frac{f(x) - f(1)}{x - 1} && \\
\\
\displaystyle m =& \lim \limits_{x \to 1} \frac{\sqrt{x} - \sqrt{1}}{x - 1}
&& \text{ Substitute value of $a$ and $x$}\\
\\
\displaystyle m =& \lim \limits_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} \cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1}
&& \text{ Multiply both numerator and denominator by $(\sqrt{x} + 1)$}\\
\\
\displaystyle m =& \lim \limits_{x \to 1} \frac{\cancel{ x - 1}}{\cancel{(x - 1)} (\sqrt{x} + 1)}
&& \text{ Cancel out like terms }\\
\\
\displaystyle m =& \lim \limits_{x \to 1} \frac{1}{\sqrt{x} + 1} = \frac{1}{\sqrt{1} + 1}
&& \text{ Evaluate the limit}\\
\\
m =& \frac{1}{2}
&& \text{ Slope of the tangent line}


\end{aligned}
\end{equation}
$


Using point slope form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1) && \\
\\
\displaystyle y - 1 =& \frac{1}{2} ( x - 1 )
&& \text{ Substitute value of $x, y$ and $m$}\\
\\
\displaystyle y =& \frac{x- 1}{2} + 1
&& \text{ Get the LCD}\\
\\
\displaystyle y =& \frac{x - 1 + 2}{2}
&& \text{ Combine like terms}\\
\\
y =& \frac{x+1}{2}
\end{aligned}
\end{equation}
$


Therefore,
The equation of the tangent line at (1,1) is $\displaystyle y = \frac{x + 1}{2}$