Odpowiedź :
Odpowiedź:
[tex]\lim_{n \to \infty} \frac{\sqrt{n^2+3}-n}{\sqrt{n^2+2}-n}= \\= \lim_{n \to \infty}\frac{(\sqrt{n^2+3}-n)(\sqrt{n^2+2}+n)(\sqrt{n^2+3}+n)}{(\sqrt{n^2+2}-n)(\sqrt{n^2+2}+n)(\sqrt{n^2+3}+n)}=\\ = \lim_{n \to \infty}\frac{(\sqrt{n^2+3}-n)(\sqrt{n^2+3}+n)(\sqrt{n^2+2}+n)}{(n^2+2-n^2)(\sqrt{n^2+3}+n)}=\\= \lim_{n \to \infty}\frac{(n^2+3-n^2)(\sqrt{n^2+2}+n)}{2(\sqrt{n^2+3}+n)}=[/tex]
[tex]= \lim_{n \to \infty} \frac{3(\sqrt{n^2+2}+n)}{2(\sqrt{n^2+3}+n)}=\\= \lim_{n \to \infty} \frac{3(\sqrt{n^2(1+\frac2{n^2})}+n)}{2(\sqrt{n^2(1+\frac3{n^2})}+n)}=\\= \lim_{n \to \infty} \frac{3(n\sqrt{1+\frac2{n^2}}+n)}{2(n\sqrt{1+\frac3{n^2}}+n)}= \\ =\lim_{n \to \infty} \frac{3n(\sqrt{1+\frac2{n^2}}+1)}{2n(\sqrt{1+\frac3{n^2}}+1)}= \\= \lim_{n \to \infty} \frac{3(\sqrt{1+\frac2{n^2}}+1)}{2(\sqrt{1+\frac3{n^2}}+1)}=\frac{3(\sqrt1+1)}{2(\sqrt1+1)} = \frac32[/tex]
[tex]\lim_{n \to \infty} \frac{2n^2-(-3^n)}{3^n+4} = \lim_{n \to \infty} \frac{2n^2+3^n}{3^n+4}= \\ = \lim_{n \to \infty}\frac{2n^2}{3^n+4} + \lim_{n \to \infty} \frac{3^n}{3^n+4}=(***)\\ \lim_{n \to \infty}\frac{2n^2}{3^n+4} = \frac{\infty}{\infty} = \lim_{n \to \infty}\frac{(2n^2)'}{(3^n+4)'} = \lim_{n \to \infty}\frac{4n}{3^n\cdot\ln3}=\\ =\frac{\infty}{\infty}= \lim_{n \to \infty}\frac{(4n)'}{(3^n\cdot\ln3)'}= \lim_{n \to \infty}\frac4{3^n\cdot\ln^23}=\frac4{\infty}=0[/tex]
[tex]\lim_{n \to \infty}\frac{3^n}{3^n+4}= \lim_{n \to \infty}\frac{3^n}{3^n(1+\frac4{3^n})}=\\ = \lim_{n \to \infty}\frac1{1+\frac4{3^n}}=\frac1{1+0} = 1 \\ (***)=0+1=1[/tex]