Question

Find the slope of the tangent line when x=3 using the limit definition f(x) = X^2 - 5

Answer 1

From the limit definition, we have that

`f^(\prime)(x)=\lim _(h\to0)(f(x+h)-f(x))/(h)`

Now applying we have

`\begin{gathered} f\mleft(x\mright)=x^2-5 \\ f^(\prime)(x)=\lim _(h\to0)(f(x+h)-f(x))/(h) \\ =\lim _(h\to0)(((x+h)^2-5)-(x^2-5))/(h) \\ =\lim _(h\to0)\frac{x^2+2xh+h^2^{}-5-(x^2-5)}{h} \\ =\lim _(h\to0)(x^2+2xh+h^2-5-x^2+5)/(h) \\ =\lim _(h\to0)(x^2-x^2+2xh+h^2-5+5)/(h) \\ =\lim _(h\to0)(2xh+h^2)/(h) \end{gathered}`

factorizing for h, we have

`\begin{gathered} =\lim _(h\to0)\frac{h(2x+h)^{}}{h} \\ \text{cancelling h} \\ =\lim _(h\to0)2x+h \\ =2x \end{gathered}`

So, when x = 3, we have

`\begin{gathered} =2x \\ =2*3 \\ =6 \end{gathered}`

Hence, the answer is 6