Question

How do you use the limit definition to find the slope of the tangent line to the graph `f(x)=sqrt(x+2)` at x=2?

Answer 1

`\frac{1}{4}`

Explanation:

By definition, the slope of the tangent line to the graph of a given function `f(x)` at a given point `x_0` is the derivative `f'(x_0)`, which in turn is defined as follows:

`f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}`

In your case, `f(x) = sqrt(x+2)` and `x_0 = 2`. The definition becomes

`f'(2) = \lim_{h \to 0} \frac{sqrt((2+h)+2)-sqrt(2+2)}{h} = \lim_{h \to 0} \frac{sqrt(4+h)-2}{h}`

To solve this limit we may "razionalize" the expression:

`\frac{sqrt(4+h)-2}{h} = \frac{sqrt(4+h)-2}{h} \cdot \frac{sqrt(4+h)+2}{sqrt(4+h)+2} = \frac{(4+h)-4}{sqrt(4+h)+2} = \frac{h}{sqrt(4+h)+2}`

The limit becomes

`f'(2) = \lim_{h \to 0}\frac{h}{h(sqrt(4+h)+2)} = \lim_{h \to 0}\frac{1}{sqrt(4+h)+2} = \frac{1}{sqrt(4+0)+2} = \frac{1}{2+2} = \frac{1}{4}`