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, $f(x)=\dfrac{\sqrt{10^2+10}}{10+1}=\dfrac{\sqrt{100+10}}{11}=0.953462589246$
Substitute $x=100$
We get, $f(x)=\dfrac{\sqrt{100^2+100}}{100+1}=\dfrac{\sqrt{10000+100}}{101}=0.99503719021$
Substitute $x=1000$
We get, $f(x)=\dfrac{\sqrt{1000^2+1000}}{1000+1}=\dfrac{\sqrt{1000000+1000}}{1001}=0.999500374688$
Substitute $x=10000$
We get, $f(x)=\dfrac{\sqrt{10000^2+10000}}{10000+1}=\dfrac{\sqrt{100000000+10000}}{10001}=0.99995000375$
Substitute $x=100000$
We get, $f(x)=\dfrac{\sqrt{100000^2+100000}}{100000+1}=\dfrac{\sqrt{10000000000+100000}}{100001}=0.999995000037$
Substitute $x=1000000$
We get, $f(x)=\dfrac{\sqrt{1000000^2+1000000}}{1000000+1}=\dfrac{\sqrt{1000000000000+1000000}}{1000001}=0.9999995$
So, we get that if $x$ increases the value of function $f(x)$ gets closer and closer to $1$.
Therefore, if $x$ increases without bound $f(x)$ should approaches $1$.
Hence, $\lim_\limits{x\to +\infty}f(x)$ should be $1$.
Or, $\lim_\limits{x\to +\infty}f(x)=1$
