Answer
$\lim_\limits{x\to -\infty}f(x)$ should be $-1$.
Or, $\lim_\limits{x\to -\infty}f(x)=-1$
Work Step by Step
We have given $f(x)=\dfrac{\sqrt{x^2+x}}{x+1}$
Substitute $x=-10$
We get, $f(x)=\dfrac{\sqrt{10^2-10}}{-10+1}=\dfrac{\sqrt{100-10}}{-9}=-1.05409255339$
Substitute $x=-100$
We get, $f(x)=\dfrac{\sqrt{100^2-100}}{-100+1}=\dfrac{\sqrt{10000-100}}{-99}=-1.00503781526$
Substitute $x=-1000$
We get, $f(x)=\dfrac{\sqrt{1000^2-1000}}{-1000+1}=\dfrac{\sqrt{1000000-1000}}{-999}=-1.00050037531$
Substitute $x=-10000$
We get, $f(x)=\dfrac{\sqrt{10000^2-10000}}{-10000+1}=\dfrac{\sqrt{100000000-10000}}{-9999}=-1.00005000375$
Substitute $x=-100000$
We get, $f(x)=\dfrac{\sqrt{100000^2-100000}}{-100000+1}=\dfrac{\sqrt{10000000000-100000}}{-99999}=-1.00000500004$
Substitute $x=-1000000$
We get, $f(x)=\dfrac{\sqrt{1000000^2-1000000}}{-1000000+1}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{\sqrt{1000000000000-1000000}}{-999999}=-1.0000005$
So, we get that if $x$ decreases the value of function $f(x)$ gets closer and closer to $-1$.
Therefore, if $x$ decreases without bound $f(x)$ should approaches $-1$.
Hence, $\lim_\limits{x\to -\infty}f(x)$ should be $-1$.
Or, $\lim_\limits{x\to -\infty}f(x)=-1$
