How to calculate $\lim\limits_{x\to5} 2\cdot \frac {x+2} {x-5}$ without L'Hopital? How to calculate  $\lim\limits_{x\to5} 2\cdot \frac {x+2} {x-5}$ without L'Hopital?




I'm making a lot of progress in learning how to calculate limits, but some still seem to be impossible for me. I don't want a concrete solution but if someone could provide a hint or a trick to progress in general it would be really helpful. I'm having a hard time here.
Direct substitution gives

$$\lim_{x\to5}2\cdot\frac{x+2}{x-5}=2\cdot\frac70\to\pm\infty$$

Can you check this reasoning?
A different approach..
$$
\frac{x+2}{x-5} = \frac{x-5 + 7}{x-5} = 1 + \frac{7}{x-5}
$$
So 
$$
\lim_{x\to 5}\frac{x+2}{x-5} =\lim_{x\to 5} 1 + \frac{7}{x-5} = 1 +\lim_{x\to 5} \frac{7}{x-5}
$$
(if this makes more sense as we only varying the denominator)