<s>[INST] Limit of $(x+1+\sqrt{(x+1)^2 +1})$ as $x\to-\infty$ This seemed easy at first, but then I confronted my result with the one I got from WolframAlpha and they are different:
$$
\lim_{x\to-\infty}(x+1+\sqrt{(x+1)^2 +1})
$$
WolframAlpha says the limit is equal to 0, though for me it seems to be negative infinity. I tried some simple algebraic manipulation, namely multiplying each factor by 1:
$$
\frac{x\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+\frac{\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+[(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}}
$$
The last part is 0, because the fraction goes to 0:
$$
\lim_{x\to-\infty}([(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}})=0
$$
The middle part is 1 and the first part should be negative infinity, as the only factor left standing is x and it goes to negative infinity. What am I missing?
 [/INST] Most of the time when you want to get rid of a root expression you use the 3rd binomial formula. So you would expand into a fraction with $(x+1-\sqrt{(x+1)^2+1})$
So we get:
$\frac{(x+1)^2-(x+1)^2-1}{x+1-\sqrt{(x+1)^2+1}}=\frac{-1}{x+1-\sqrt{(x+1)^2+1}}=\frac{1}{\sqrt{(x+1)^2+1}-(x+1)}$
We have $\sqrt{(x+1)^2+1}-(x+1)\to\infty$ as $x\to -\infty$. So the fraction goes to $0$.
</s>