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$\lim_{x \rightarrow \infty} \sqrt{x^2+7x}-x$

I'm having trouble finding the problem with the method I'm using:

$\lim_{x \rightarrow \infty} \sqrt{x^2(1+\frac{7}{x})}-x$

$\lim_{x \rightarrow \infty} x\sqrt{1+\frac{7}{x}}-x$

$\lim_{x \rightarrow \infty} \frac{7}{x}=0$

$\lim_{x \rightarrow \infty} x\sqrt{1+0}-x$

so I get 0 for my answer. I know that graphing the function makes it clear that this is not the case but I don't see whats wrong with what I did above.

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Hint: When you have that $$\lim_{x \rightarrow \infty} f(x) = \infty$$ and $$\lim_{x \rightarrow \infty} g(x) = \infty,$$ can you then conclude that $$\lim_{x \rightarrow \infty} [f(x)-g(x)] = 0?$$

(Consider $f(x) = 2x$ and $g(x) = x$.)

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Idea to solve the problem using the expression's conjugate:

$$\left(\sqrt{x^2+7x}-x\right)\cdot\frac{\left(\sqrt{x^2+7x}+x\right)}{\left(\sqrt{x^2+7x}+x\right)}=\frac{x^2+7x-x^2}{\sqrt{x^2+7x}+x}=$$

$$=\frac 7{\sqrt{1+\frac7x}+1}\xrightarrow[x\to\infty]{}\frac72$$

The problems with what you did: (1) You took the limit only of part of the expression ($\frac7x$) while leaving the rest untouched, and this is strictily forbidden unless you prove otherwise in your particular case, and (2) even after this, you remain with an expression that doesn't help you solve the problem as you get " x time something minus x"... and that can be anything.

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