$$lim_{x\to-0.2^+} \sqrt{\frac{x+11}{x+2}}$$
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: $$\sqrt[2]{\frac{\lim_{x\to-0.2^+}x+11}{\lim_{x\to-0.2^+}x+2}}$$ 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?
[Math] Finding the limit by using roots, quotient and sum laws.
calculuslimits
Best Answer
Moving my comment(s) to the answer zone.
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.