Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1155: 81

The value of $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$ for $ f\left( x \right)=\sqrt{x}$, $ a=1$ is $\frac{1}{2}$.

The function given is $ f\left( x \right)=\sqrt{x}$. As $ a=1$, $\begin{align} & f\left( a+h \right)=f\left( 1+h \right) \\ & =\sqrt{1+h} \end{align}$ $\begin{align} & f\left( a \right)=f\left( 1 \right) \\ & =\sqrt{1} \\ & =1 \end{align}$ Consider the provided limit $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$. Substitute the values of $ f\left( a+h \right)$ and $ f\left( a \right)$: $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+h}-1}{h}$ Rationalize the numerator by multiplying the numerator and denominator by $\sqrt{1+h}+1$: $\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+h}-1}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+h}-1}{h}\times \frac{\sqrt{1+h}+1}{\sqrt{1+h}+1} \end{align}$ Using $\left( \sqrt{a}-\sqrt{b} \right)\left( \sqrt{a}+\sqrt{b} \right)={{\left( \sqrt{a} \right)}^{2}}-{{\left( \sqrt{b} \right)}^{2}}=a-b $: $\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+h}-1}{h}\times \frac{\sqrt{1+h}+1}{\sqrt{1+h}+1} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{1+h-1}{h\left( \sqrt{1+h}+1 \right)} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{h\left( \sqrt{1+h}+1 \right)} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{1+h}+1} \end{align}$ Using the quotient property of limits $\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\frac{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)},\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)\ne 0$: $\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{1+h}+1} \\ & =\frac{\underset{h\to 0}{\mathop{\lim }}\,1}{\underset{h\to 0}{\mathop{\lim }}\,\left( \sqrt{1+h}+1 \right)} \end{align}$ Use limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ in the denominator: $\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{\underset{h\to 0}{\mathop{\lim }}\,1}{\underset{h\to 0}{\mathop{\lim }}\,\left( \sqrt{1+h}+1 \right)} \\ & =\frac{\underset{h\to 0}{\mathop{\lim }}\,1}{\underset{h\to 0}{\mathop{\lim }}\,\sqrt{1+h}+\underset{h\to 0}{\mathop{\lim }}\,1} \end{align}$ Using limit property $\underset{x\to a}{\mathop{\lim }}\,\sqrt{f\left( x \right)}=\sqrt{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}$: $\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{\underset{h\to 0}{\mathop{\lim }}\,1}{\underset{h\to 0}{\mathop{\lim }}\,\sqrt{1+h}+\underset{h\to 0}{\mathop{\lim }}\,1} \\ & =\frac{\underset{h\to 0}{\mathop{\lim }}\,1}{\sqrt{\underset{h\to 0}{\mathop{\lim }}\,\left( 1+h \right)}+\underset{h\to 0}{\mathop{\lim }}\,1} \end{align}$ Use limit property $\underset{x\to a}{\mathop{\lim }}\,c=c $: $\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\frac{\underset{h\to 0}{\mathop{\lim }}\,1}{\sqrt{\underset{h\to 0}{\mathop{\lim }}\,\left( 1+h \right)}+\underset{h\to 0}{\mathop{\lim }}\,1} \\ & =\frac{1}{\sqrt{1+0}+1} \\ & =\frac{1}{\sqrt{1}+1} \\ & =\frac{1}{2} \end{align}$