<s>[INST] Rule for factoring out variable in limit I am a bit confused about the proper way to calculate the limit of a function by factoring out the variable that strives for infinity.
If you have this:
$\lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x+1}-\sqrt{x}\right)$
for some reason you cannot do this:
$ = \lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x(1+ \dfrac{1}{x})}-\sqrt{x}\right) = \lim_{x\rightarrow \infty } \sqrt{x} \left(\sqrt{x}-\sqrt{x}\right) = 0$
but rather should do this:
$ = \lim_{x\rightarrow \infty } \sqrt{x} \frac{\left(\sqrt{x+1}-\sqrt{x}\right)\left(\sqrt{x+1}+\sqrt{x}\right)}{\left(\sqrt{x+1}+\sqrt{x}\right)} = \lim_{x\rightarrow \infty } \sqrt{x} \frac{x+1-x}{\left(\sqrt{x+1}+\sqrt{x}\right)} = \lim_{x\rightarrow \infty } \frac{\sqrt{x}}{\sqrt{x(1+\frac{1}{x})}+\sqrt{x}} = \lim_{x\rightarrow \infty } \frac{\sqrt{x}}{2\sqrt{x}} = \frac{1}{2}$
Why is it legit to factor out $x$ in the second approach but not in the first one? Is there a general rule for when you are allowed to do so and when not?
 [/INST] Although $\lim_{x \to \infty} \sqrt{x} (3-3) = \lim_{x \to \infty} \sqrt{x} \cdot 0 =0$, in your case you have $\lim_{x \to \infty} = \sqrt{x}(\sqrt{x} - \sqrt{x}) \neq \lim_x \sqrt{x} \cdot (\lim_x \sqrt{x} - \sqrt{x})$ because $\infty \cdot 0 $ is an indeterminate form. Also $\lim_x (\sqrt{x} - \sqrt{x}) \neq \lim_x \sqrt{x} - \lim_x \sqrt{x}$ because \infty - \infty is an indeterminate form too.
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