Why $\lim_{x \to \infty } \sqrt{(x^2 + 7x)} - x$ is $7\over2$ I know how to get the correct answer. 

$$\lim_{x \to \infty } \sqrt{(x^2 + 7x)} - x$$
  $$\lim_{x \to \infty } \frac{(x^2 + 7x) - x^2}{\sqrt{(x^2 + 7x)} + x}$$
  $$\lim_{x \to \infty } \frac{7x}{\sqrt{x(x + 7)} + x}$$
  $$\lim_{x \to \infty } \frac{7 \sqrt{x}}{\sqrt{(x + 7)} + \sqrt{x}}$$

Now as $x \to \infty$ the value of $\lim_{x \to \infty } \sqrt{(x + 7)}$ becomes similar to $\lim_{x \to \infty } \sqrt{x}$

$$\lim_{x \to \infty } \frac{7 \sqrt{x}}{\sqrt{x} + \sqrt{x}} = \frac{7}{2}$$

I want to know why this other way is incorrect :-

$$\lim_{x \to \infty } \sqrt{(x^2 + 7x)} - x$$

Now as $x \to \infty$ the value of $\lim_{x \to \infty } \sqrt{(x^2 + 7x)}$ becomes similar to $\lim_{x \to \infty } \sqrt{x^2}$

$$\lim_{x \to \infty } \sqrt{(x^2)} - x = 0/-2$$

Which is incorrect, i checked on a graphing calculator. But i want to know why ? please help. 
 A: We can say that they are almost the same, so that in the difference the equal part cancel and what remains is exactly what in which they differ, so neglecting this part leads to an error.
A: Hint $\infty-\infty$ is an indeterminate form it's not equal to $0$ so to get some finite value we multiply by conjugate 
A: Let $f:x \mapsto \sqrt x$ and $g: x \mapsto x^2+7x$.
So you noticed that $g \sim (x\mapsto x^2)$ and then you concluded from that that $f\circ g \sim (x \mapsto f(x^2))$. Be this equivalence relation is generally not compatible with composition, hence the wrong result.
A: I think students often think that limits are an approximate tool and when dealing with limits we can replace $A$ by $B$ even when $A \neq B$. Sorry!!! This is never the case in mathematics. You can replace $A$ by $B$ only when $A = B$.
Thus in your first approach when you reach the fraction $$\frac{7\sqrt{x}}{\sqrt{x + 7} + \sqrt{x}}$$ then you are not allowed to replace $\sqrt{x + 7}$ by $\sqrt{x}$ by saying that they are similar. What you can and should do is to divide both numerator and denominator by $\sqrt{x} \neq 0$ to get the fraction $$\dfrac{7}{\sqrt{1 + \dfrac{7}{x}} + 1}$$ and then we can see that $7/x \to 0$ as $x \to \infty$ so the above fraction tends to $7/(\sqrt{1 + 0} + 1) = 7/2$.

You have yourself shown the danger of replacing $A$ by $B$ when they are similar but not equal in your second evaluation. A wrong approach may give a right answer sometimes but that does not make it a right approach.
A: In the second way, you are making a mistake writing

the value of $\lim_{x \to \infty } \sqrt{(x + 7)}$    becomes similar to $\lim_{x \to \infty } \sqrt{x}$

In fact, for large values of $x$,
$$\sqrt{x^2 + 7x}=x \sqrt{1+\frac 7 x}\approx x\left(1+\frac 7 {2x} \right)$$
