Why is it required to change variable to get the right answer for this question? The question is this :
$$\lim_{x\to-\infty} {\sqrt{x^2+x}+\cos x\over x+\sin x}$$
The solution is $-1$ and this seems to be only obtained from the change variable strategy, such as $t=-x$.
However, I have no idea why this isn't just solved by simply eliminating $x$ in numerator and denominator, which generates the value $1$.
It seems that this is related with $x\to-\infty$, but I have no specific idea.
Can anyone help me? 
 A: Note that the trigonometric terms are negligible as $x \to-\infty $. Hence,
$$\lim_{x\to-\infty}\frac{\sqrt{x^2+x}}{x}$$
You cannot take the x in the square root. But, your problem is that x is negative. So, you must : 
$$\lim_{x\to-\infty}\frac{\sqrt{x^2+x}}{-(-x)}$$
Now, $-x$ is positive and you can take it inside the root where it will become squared. 
$$\lim_{x\to-\infty}\frac{\sqrt{1+1/x}}{-1}$$
Clearly, the answer is $-1$.
Substituing $t=-x$, just makes it look good. Note that you can write $-x$ as $t$ every time in your substituted answer.
A: $t = -x$ gives: $L$ = $-\displaystyle \lim_{t \to \infty} \dfrac{\sqrt{t^2 - t} + \cos t}{t + \sin t} = -\displaystyle \lim_{t \to \infty} \dfrac{\sqrt{1 - \dfrac{1}{t}} + \dfrac{\cos t}{t}}{1 + \dfrac{\sin t}{t}} = -1$ because $\displaystyle \lim_{t \to \infty} \dfrac{\cos t}{t}  = \displaystyle \lim_{t \to \infty} \dfrac{\sin t}{t} = \displaystyle \lim_{t \to \infty} \dfrac{1}{t} = 0$
A: You are thinking to do $\sqrt{x^2 + x} = x\sqrt{1+\frac 1 {x^2}}$, however, $\sqrt{x^2 + x}$ and $\sqrt{1+\frac 1 {x^2}}$ are (defined to be) positive, so how can you do this with a negative $x$?
