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?
 A: 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 $0$.
A: I don't like limits at $-\infty$. I'd rewrite it as
$$\lim_{x\to\infty}\left(1-x+\sqrt{1+(x-1)^2}\right)
=\lim_{y\to\infty}\left(-y+\sqrt{1+y^2}\right)$$
(setting $y=x-1$). Then
$$-y+\sqrt{1+y^2}=\frac{-y^2+(1+y^2)}{y+\sqrt{1+y^2}}
=\frac{1}{y+\sqrt{1+y^2}}\to0$$
as $y\to\infty$.
A: Your mistake was in writing $\lim_{x\to -\infty}\frac{(x+1)^2+1}{\sqrt{(x+1)^2+1}}=0$. The expression $\frac{(x+1)^2+1}{\sqrt{(x+1)^2+1}}$ equals $\sqrt{(x+1)^2+1}$, which tends to $\infty$ as $x\rightarrow-\infty$.
You can attack this problem using other techniques. One approach is to multiply and divide by the conjugate of $x+1+\sqrt{(x+1)^2+1}$, namely $x+1-\sqrt{(x+1)^2+1}$, then use the difference of squares formula to simplify the result. Alternatively, if you are comfortable with hyperbolic functions and the exponential function $e^x$, you could write
$$x+1+\sqrt{(x+1)^2+1}=e^{\sinh ^{-1}(x+1)}$$
and use the fact that $\lim_{x\to -\infty}\sinh ^{-1}(x)=-\infty$ (this limit can be proven without using the logarithmic form of $\sinh ^{-1}x$). Since $\lim_{x\to -\infty}e^x=0$, we have that
$$\lim_{x\to -\infty}e^{\sinh ^{-1}(x+1)}=0$$
Thus,
$$\lim_{x\to -\infty}\left(x+1+\sqrt{(x+1)^2+1}\right)=0$$.
A: $$
\lim_{x\to-\infty}\left(x+1+\sqrt{(x+1)^2 +1}\right)
$$
$$=\lim_{x\to-\infty}\left(x+1+|x+1|\sqrt{1+\dfrac1{(x+1)^2}}\right)$$
$$=\lim_{x\to-\infty}\left(x+1+|x+1|\left[1+\frac1{2(x+1)^2}\cdots\right]\right)$$
$$=\lim_{x\to-\infty}\left(x+1-(x+1)-(x+1)\left[\frac1{2(x+1)^2}\cdots\right]\right)=0.$$
