Limit Problem of square root equation

Question

i have tried to solve the attached limit problem(18th) and got the result as 1/4. however, the book says it has to be 1/16. can anyone explain me why please? regards,

Answer 1

Using the limit as it seems to be indicated by your question,

The slope $\alpha$ is given by $$\alpha=\lim_{x \rightarrow 4}\frac{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{4}}}{x-4}$$

$$\frac{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{4}}}{x-4}=\frac{\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{4}}\right) \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{4}}\right)}{\left(x-4\right) \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{4}}\right)}=\frac{\frac{1}{x}-{\frac{1}{4}}}{\left(x-4\right) \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{4}}\right)}=\frac{\frac{4-x}{4x}}{\left(x-4\right) \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{4}}\right)}=-\frac{1}{4x\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{4}}\right)}$$

Replacing $x$ with $4$, you get $-\frac{1}{16}$