Answer
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)={{x}^{2}}+2x-3$, $ a=1$ is 4.
Work Step by Step
The function is given as
$ f\left( x \right)={{x}^{2}}+2x-3$
As $ a=1$,
$\begin{align}
& f\left( a+h \right)=f\left( 1+h \right) \\
& ={{\left( 1+h \right)}^{2}}+2\left( 1+h \right)-3 \\
& ={{1}^{2}}+\left( 2\times 1\times h \right)+{{h}^{2}}+2+2h-3 \\
& =1+2h+{{h}^{2}}+2+2h-3
\end{align}$
Solve further:
$ f\left( a+h \right)={{h}^{2}}+4h $
$\begin{align}
& f\left( a \right)=f\left( 1 \right) \\
& ={{1}^{2}}+\left( 2\times 1 \right)-3 \\
& =1+2-3 \\
& =0
\end{align}$
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)$
$\begin{align}
& \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}+4h-0}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}+4h}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( h+4 \right)}{h}
\end{align}$
Cancel out the common factor $ h $ from both the numerator and denominator.
$\begin{align}
& \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( h+4 \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\left( h+4 \right)
\end{align}$
Using limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$
$\begin{align}
& \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\left( h+4 \right) \\
& =\underset{h\to 0}{\mathop{\lim }}\,h+\underset{h\to 0}{\mathop{\lim }}\,4
\end{align}$
Use the property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{ }$ in $\underset{h\to 0}{\mathop{\lim }}\,4$:
$\begin{align}
& \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=0+4 \\
& =4
\end{align}$
Thus, 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)={{x}^{2}}+2x-3$, $ a=1$ is 4.
