# How to obtain: $\lim_{x \to -\infty}\big(\sqrt{x^2+3x+2}+x-1\big)$?

Please help me to do my homework Take look to the below image I would like a solution step by step of $$\lim_{x \to -\infty}\left(\sqrt{x^2+3x+2}+x-1\right).$$ enter image description here

• See the method : math.stackexchange.com/questions/1766582/… – lab bhattacharjee Dec 4 '16 at 17:45
• "Please help me to do my homework" - well, at least you are honest about it. But it's not recommended to use Math.SE as a homework help site, especially without any effort of your own shown. – Yuriy S Dec 4 '16 at 18:38
• @YuriyS Yes I did effort, but when I try to draw a curve I find it illogical. I know where I do the error but I do not have enough skills. When I put the limit in [link](wolframalpha.com) I got a result -5/2 . but I do not understand how to count it because I forgot when you drop some variable out of the root, you must add the absolute value. That's it and sorry if I did some linguistic errors. – Mourad Karoudi Dec 4 '16 at 19:01

Hint. One may write, as $x \to -\infty$, \begin{align} \sqrt{x^2+3x+2}+x-1&=\frac{(x^2+3x+2)-(x-1)^2}{\sqrt{x^2+3x+2}-(x-1)} \\\\&=\frac{5+1/x}{-\sqrt{1+3/x+2/x^2}-1+1/x} \end{align} where we have used $\sqrt{x^2}=|x|=-x$ since $x<0$.
• @Mourad Kaaroudi You are welcome. Do you get this: $\sqrt{x^2}=|x|=-x$ for $x<0$? Recall that for example the absolute value of $-7$ is $|-7|=-(-7)=7$ since $-7<0$. – Olivier Oloa Dec 4 '16 at 17:35
Similarly completing the square of $x^2+3x+2=(x+\frac{3}{2})^2-\frac{1}{4}$ as we go to $-\infty$ we get that the square root of this is equivalent to $-x-\frac{3}{2}$ because the $-\frac{1}{4}$ isn't going to matter as we go to $-\infty$ now our limit is $\lim_{x\to-\infty} -x-\frac{3}{2}+x-1$ which is then just $-\frac{5}{2}$