Question

use the limit definition of the derivative to find the slope of the tangent line to the curve $f(x)=5x^{2}$ at $x = 2$.
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Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ at $x=a$ is $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $f(x)=5x^{2}$ and $a = 2$. First, find $f(2 + h)$ and $f(2)$.
$f(2 + h)=5(2 + h)^{2}=5(4 + 4h+h^{2})=20 + 20h+5h^{2}$, and $f(2)=5\times2^{2}=20$.

Step2: Substitute into the limit - formula

$f^{\prime}(2)=\lim_{h
ightarrow0}\frac{f(2 + h)-f(2)}{h}=\lim_{h
ightarrow0}\frac{(20 + 20h+5h^{2})-20}{h}$.
Simplify the numerator: $\frac{(20 + 20h+5h^{2})-20}{h}=\frac{20h+5h^{2}}{h}$.

Step3: Simplify the fraction

Since $h
eq0$ (as we are taking the limit as $h$ approaches 0, not setting $h = 0$), we can cancel out the $h$ in the numerator and denominator. $\frac{20h+5h^{2}}{h}=\frac{h(20 + 5h)}{h}=20 + 5h$.

Step4: Evaluate the limit

Now, find $\lim_{h
ightarrow0}(20 + 5h)$. As $h$ approaches 0, we substitute $h = 0$ into $20+5h$. $\lim_{h
ightarrow0}(20 + 5h)=20$.

Answer:

$20$