# Finding the limit by using roots, quotient and sum laws.

$$lim_{x\to-0.2^+} \sqrt{\frac{x+11}{x+2}}$$ I have tried working this problem out and the limit I got was 2.44. My first attempt was just plugging in $-0.2$ and solving for the limit. I was told to first use the root rule of a limit law, then the quotient rule and sum rule and simplify. I have tried to do this but I am now stuck here: $$\sqrt{\frac{\lim_{x\to-0.2^+}x+11}{\lim_{x\to-0.2^+}x+2}}$$ Is this the correct approach to the problem? If I were to simplify this I end up with the same answer as I would if I plugged in $-0.2$. What am I doing wrong here?

• The function $f(x)$ is continuous at $-0.2$, so the limit is $f(-0.2)$. Your first attempt was the right one. The second approach is fine too. And leads to the same answer. – Julien Feb 12 '13 at 2:43
• I have tried 2.44 as an answer but the website said it was incorrect. – Kot Feb 12 '13 at 2:46
• $2.44$ is an approximation. Try the exact value with square roots? – Julien Feb 12 '13 at 2:48
• That was it! Thank you :D. – Kot Feb 12 '13 at 2:52
• @JacobBlack Thanks for your you comment. Will do. Thanks again. – Julien Feb 12 '13 at 3:05

The function $$f(x)=\sqrt{\frac{x+11}{x+2}}$$ is continuous on its domain $(-\infty,-11]\cup(-2,+\infty)$ by composition of well-known continuous functions.
Since $-0.2$ belongs to the domain above (note it is in the interior, so the following limit can be taken from the left and the right simultaneously), we have, by definition of continuity: $$\lim_{x\rightarrow -0.2} f(x)=f(-0.2)=\sqrt{\frac{10.8}{1.8}}=\sqrt{\frac{108}{18}}=\sqrt{6}.$$
Note that the right limit you are asking about is a fortiori $\sqrt{6}$.
Finally, the second approach is fine too and leads to the same answer. But I tend to prefer the one above, since continuity at $-0.2$ is really what this is about.