Desarrollo de algunos Límites.
1.- $\displaystyle \lim_{x\to 2} \frac{x^2+1}{x^2-1}=\frac{2^2+1}{2^2-1}=\frac{5}{3}$
$$\displaystyle \lim_{x\to 2} \frac{x^2+1}{x^2-1}=\frac{2^2+1}{2^2-1}=\frac{5}{3}$$
2.- $\displaystyle \lim_{x\to 1} \frac{x^2-2x+1}{x^3-x}=\frac{1^2-2\cdot 1+1}{1^3-1}=\frac{0}{0}$
$$\displaystyle \lim_{x\to 1} \frac{x^2-2x+1}{x^3-x}=\frac{1^2-2\cdot 1+1}{1^3-1}=\frac{0}{0}$$
pero
$$\displaystyle \lim_{x\to 1} \frac{x^2-2x+1}{x^3-x}=\lim_{x\to 1}\frac{(x-1)^2}{x(x^2-1)}$$ $$= \\ \lim_{x\to 1}\frac{(x-1)^2}{x(x-1)(x+1)}=\lim_{x\to 1}\frac{x-1}{x(x+1)}=\frac{1-1}{1(2)}=\frac{0}{2}=0$$
3.-$\displaystyle \lim_{x\to 1} \frac{(x-1)\sqrt{2-x}}{1-x^2}$
$$\displaystyle \lim_{x\to 1} \frac{(x-1)\sqrt{2-x}}{1-x^2}=\frac{(1-1)\sqrt{2-1}}{1-1^2}=\frac{0\sqrt{1}}{0}=\frac{0}{0}$$
pero
$$\displaystyle \lim_{x\to 1} \frac{(x-1)\sqrt{2-x}}{(1-x)(1+x)}= \lim_{x\to 1} \frac{-(1-x)\sqrt{2-x}}{(1-x)(1+x)}=\lim_{x\to 1} \frac{-\sqrt{2-x}}{(1+x)}=\frac{-1}{2}$$
4.- $\displaystyle \lim_{x\to 1} \frac{1}{1-x}-\frac{3}{1-x^3}$
$$\displaystyle \lim_{x\to 1} \frac{1}{1-x}-\frac{3}{1-x^3}=\frac{1}{1-1}-\frac{3}{1-1^3}=\frac{1}{0}-\frac{3}{0}= \infty-\infty$$
Obviamente un indeterminado, pero
$$\displaystyle \lim_{x\to 1} \frac{1}{1-x}-\frac{3}{1-x^3}=\lim_{x\to 1} \frac{1}{1-x}-\frac{3}{(1-x)(1+x+x^2)} $$ $$=\lim_{x\to 1} \frac{1+x+x^2}{(1-x)(1+x+x^2)}-\frac{3}{(1-x)(1+x+x^2)} $$ $$ = \lim_{x\to 1} \frac{1+x+x^2-3}{(1-x)(1+x+x^2)} \\= \lim_{x\to 1} \frac{x^2+x-2}{(1-x)(1+x+x^2)}$$ $$= \lim_{x\to 1} \frac{(x-1)(x+2)}{(1-x)(1+x+x^2)}$$ $$= \lim_{x\to 1} \frac{-(1-x)(x+2)}{(1-x)(1+x+x^2)}$$ $$= \lim_{x\to 1} \frac{-(x+2)}{1+x+x^2}=\frac{-3}{3}=-1$$
5.- $\displaystyle \lim_{x\to 1}\frac{x+2}{x^2-5x+4}+\frac{x-4}{3(x^2-3x+2)}$
$$\displaystyle \lim_{x\to 1}\frac{x+2}{x^2-5x+4}+\frac{x-4}{3(x^2-3x+2)}=\frac{1+2}{1-5+4}+\frac{1-4}{3(1-3+2)}=\frac{1}{0}-\frac{1}{0}= \infty-\infty$$
también indeterminado, pero
$$\displaystyle \lim_{x\to 1}\frac{x+2}{x^2-5x+4}+\frac{x-4}{3(x^2-3x+2)}=\lim_{x\to 1}\frac{x+2}{(x-4)(x-1)}+\frac{x-4}{3(x-2)(x-1)}$$ $$=\lim_{x\to 1}\frac{3(x+2)(x-2)+(x-4)(x-4)}{3(x-4)(x-2)(x-1)}$$ $$= \lim_{x\to 1}\frac{3(x^2-4)+(x-4)^2}{3(x-4)(x-2)(x-1)}$$ $$= \lim_{x\to 1}\frac{3(x^2-4)+(x-4)^2}{3(x-4)(x-2)(x-1)}$$ $$= \lim_{x\to 1}\frac{3x^2-12+x^2-8x+16}{3(x-4)(x-2)(x-1)}$$ $$=\lim_{x\to 1}\frac{4x^2-8x+4}{3(x-4)(x-2)(x-1)}$$ $$= \lim_{x\to 1}\frac{4(x^2+1)}{3(x-4)(x-2)(x-1)}$$ $$=\lim_{x\to 1}\frac{4(x-1)^2}{3(x-4)(x-2)(x-1)}$$ $$=\lim_{x\to 1}\frac{4(x-1)}{3(x-4)(x-2)}=\frac{0}{24}=0$$
En desarrollo.
6.- $\displaystyle \lim_{x\to \infty}{2+x-3x^3}{5-x^2+3x^3}$$
7.- $\displaystyle \lim_{x\to \infty}\frac {x^3}{x^2+1}$
8.- $\displaystyle \lim_{x\to \infty}\frac {\sqrt{x^2+1}-3\sqrt{x}}{\sqrt[4]{x^3+5x-x}-x}$
9.- $\displaystyle \lim_{x\to 0}\frac {\sqrt{1+x^2}-1}{x}$
10.-$\displaystyle \lim_{x\to 5}\frac {\sqrt{x-1}-2}{x-5}$
11.- $\displaystyle \lim_{x\to 1}\frac {\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}$
12.- $\displaystyle \lim_{x\to \infty} \sqrt{(x+m)(x+n)}-x$
13.- $\displaystyle \lim_{x\to 2}\frac{1}{2-x}-\frac{3}{8-x^3}$
14.- $\displaystyle \lim_{x\to \frac {1}{2}}\frac {8x^2-1}{6x^2-5x+1}$
15.- $\displaystyle \lim_{x\to \infty}\frac {3x^4-2x+1}{3x^2+6x-2}$
16.- $\displaystyle \lim_{x\to 0}\frac {\sqrt{x^2+4}-2}{\sqrt{x^2+9}-3}$
17.- $\displaystyle \lim_{x\to 4}\frac {\sqrt{1+2x}-3}{x-4}$
18.- $\displaystyle \lim_{x\to \infty}\frac{3^x+x}{3^x-2x}$
19.- $\displaystyle \lim_{x\to \infty}\frac {\sqrt[3]{x^2+1}}{x-2}$
20.- $\displaystyle \lim_{x\to \infty}\frac{2x^2+1}{\sqrt[3]{x^7+2x}}$
21.- $\displaystyle \lim_{x\to 0}\frac {\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$
22.- $\displaystyle \lim_{x\to 1}\frac{1-\sqrt[3]{x}}{1-\sqrt{x}}$
