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Question Number 152441 by fotosy2k last updated on 28/Aug/21

Commented by fotosy2k last updated on 28/Aug/21

$$plshelp$$

Answered by qaz last updated on 28/Aug/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\sqrt{{4\mathrm{n}}^2+\mathrm{n}}-2\mathrm{n})$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{(\sqrt{{4\mathrm{n}}^2+\mathrm{n}}-2\mathrm{n})(\sqrt{{4\mathrm{n}}^2+\mathrm{n}}+2\mathrm{n})}{\sqrt{{4\mathrm{n}}^2+\mathrm{n}}+2\mathrm{n}}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\mathrm{n}}{\sqrt{{4\mathrm{n}}^2+\mathrm{n}}+2\mathrm{n}}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\sqrt{4+\frac{1}{\mathrm{n}}}+2}$$$$=\frac{1}{4}$$

Commented by fotosy2k last updated on 30/Aug/21

$$thankyou$$

Answered by puissant last updated on 28/Aug/21

$$={lim}_{n\to \mathrm{\infty }}\frac{4n^2+n-4n^2}{\sqrt{4n^2+n}+2n}$$$$={lim}_{n\to \mathrm{\infty }}\frac{n}{n(\sqrt{4+\frac{1}{n}}+2)}=\frac{1}{2+2}=\frac{1}{4}$$$$B)$$

Commented by fotosy2k last updated on 30/Aug/21

$$thankyou$$

Answered by EDWIN88 last updated on 28/Aug/21

$$\underset{n\to \mathrm{\infty }}{\lim }\sqrt{4n^2+n}-2n=\underset{n\to \mathrm{\infty }}{\lim 2}(n+\frac{1}{2\times 4})-2n$$$$=\frac{1}{4}$$

Commented by fotosy2k last updated on 30/Aug/21

$$thankyou$$

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