Background : I'm currently studying for my calc mid-term which is coming up, and stumbled upon a couple of questions which I don't understand on how to solve.
Question : Evaluate $$\lim_{x\to-4} \frac{\large \frac{1}{4} + \frac{1}{x}}{4 + x}$$
My Thought Process : I realize that just "plugging in" won't work, so I need to multiply by the conjugate $\large \frac{1}{4} + \frac{1}{x}$, however that's where my problem comes in. I don't remember, how to do this.
Also, I have a similar problem but with a root.
$$\lim_{x\to 7} \frac{\sqrt{x+2}-3}{x-7}$$
Once again, plugging in won't work, so I need to multiply by it's conjugate $\sqrt{x+2} + 3$, however I don't know what the result would be. If anyone could help explain this to me I would greatly appreciate it.
Here are couple of observations, which should enable you to get your answer.
For the first one, $$\dfrac{\dfrac14 + \dfrac1x}{4+x} = \dfrac{\dfrac{x+4}{4x}}{4+x} = \dfrac1{4x} \,\,\,\,\,\, \text{for }x \neq -4$$
For the second one, \begin{align} \dfrac{\sqrt{x+2}-3}{x-7} & = \dfrac{\sqrt{x+2}-3}{x-7} \times \dfrac{\sqrt{x+2}+3}{\sqrt{x+2}+3} = \dfrac{(\sqrt{x+2})^2 - 3^2}{(x-7)(\sqrt{x+2} + 3)} = \dfrac{x+2-9}{(x-7)(\sqrt{x+2}+3)}\\ & = \dfrac{x-7}{(x-7)(\sqrt{x+2} + 3)} = \dfrac1{\sqrt{x+2}+3} \,\,\,\,\,\, \text{for }x \neq 7 \end{align}