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 $\displaystyle y = \frac{x + 1}{2}$