Question

How to evaluate question number 4

Answer 1

There are 2 ways that spring to mind.

One is to use the definition of the derivative at a point:

`f'(c)=\displaystyle\lim_(x\to c)(f(x)-f(c))/(x-c)`

In this case,
`c=0` and
`f(x)=√(5-x)`. Then

`f'(x)=-\frac1{2√(5-x)}\implies f'(0)=\displaystyle\lim_(x\to0)\frac{√(5-x)-\sqrt5}x=-\frac1{2\sqrt5}`

- - -

If you don't know about derivative yet (I would think you do, considering this is for a midterm in AP Calc, but I digress), the other way is to rely on algebraic manipulation. Multiply the numerator and denominator by the conjugate of the numerator to get

`\frac{√(5-x)-\sqrt5}x\cdot(√(5-x)+\sqrt5)/(√(5-x)+\sqrt5)=(\left(√(5-x)\right)^2-\left(\sqrt5\right)^2)/(x\left(√(5-x)+\sqrt5\right))=-\frac x{x\left(√(5-x)+\sqrt5\right)}=-\frac1{√(5-x)+\sqrt5}`

This is continuous at
`x=0`, so the limit is the value of the expression at
`x=0`:

`\displaystyle\lim_(x\to0)\frac{√(5-x)-\sqrt5}x=-\lim_(x\to0)\frac1{√(5-x)+\sqrt5}=-\frac1{2\sqrt5}`