<s>[INST] Convergence Proof: $\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$ I have to check whether the following expression converges; if yes I have to give the limit.
$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$
Now I did the following:
$$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$$
$$\lim_{x\rightarrow\infty} x\sqrt{\frac{4}{x}+1}- x\sqrt{1+\frac{1}{x}}$$
$$\lim_{x\rightarrow\infty} x \lim_{x\rightarrow\infty}(\sqrt{\frac{4}{x}+1}- \sqrt{1+\frac{1}{x}})$$
That confuses me. The left limit approaches $\infty$ while the right approaches $0$. What is wrong here or what can I conclude from that?
Thank you very muvh for your help in advance!
FunkyPeanut
 [/INST] $\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x} = \lim_{x\rightarrow\infty} \frac{(\sqrt{4x+x^2}- \sqrt{x^2+x})\cdot(\sqrt{4x+x^2}+ \sqrt{x^2+x})}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{4x+x^2-x^2-x}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}}  = \lim_{x\rightarrow\infty} \frac{3x}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{3}{ \sqrt{\frac{4}{x}+1}+ \sqrt{1+\frac{1}{x}}} = \frac{3}{ \lim_{x\rightarrow\infty}\sqrt{\frac{4}{x}+1}+ \sqrt{1+\frac{1}{x}}} = \frac{3}{\sqrt{1}+\sqrt{1}} =\frac{3}{2}$
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