Desarrollo de algunos Límites

  Cálculo

Desarrollo de algunos Límites.


1.- $\displaystyle \lim_{h\to 2} \frac{x^2+1}{x^2-1}=\frac{2x^2+1}{2x^2-1}=\frac{5}{3}$

$$\displaystyle \lim_{h\to 2} \frac{x^2+1}{x^2-1}=\frac{2x^2+1}{2x^2-1}=\frac{5}{3}$$


 

2.- $\displaystyle  \lim_{h\to 1} \frac{x^2-2x+1}{x^3-x}=\frac{1^2-2\cdot 1+1}{1^3-1}=\frac{0}{0}$

$$\displaystyle \lim_{h\to 1} \frac{x^2-2x+1}{x^3-x}=\frac{1^2-2\cdot 1+1}{1^3-1}=\frac{0}{0}$$

pero

$$\displaystyle \lim_{h\to 1} \frac{x^2-2x+1}{x^3-x}=\lim_{h\to 1}\frac{(x-1)^2}{x(x^2-1)}=\lim_{h\to 1}\frac{(x-1)^2}{x(x-1)(x+1)}=\lim_{h\to 1}\frac{x-1}{x(x+1)}=\frac{1-1}{1(2)}=\frac{0}{2}=0$$


 

3.-$\displaystyle \lim_{h\to 1} \frac{(x-1)\sqrt{2-x}}{1-x^2}$

$$\displaystyle \lim_{h\to 1} \frac{(x-1)\sqrt{2-x}}{1-x^2}=\frac{(1-1)\sqrt{2-1}}{1-1^2}=\frac{0\sqrt{1}}{0}=\frac{0}{0}$$

pero

$$\displaystyle \lim_{h\to 1} \frac{(x-1)\sqrt{2-x}}{(1-x)(1+x)}= \lim_{h\to 1} \frac{-(1-x)\sqrt{2-x}}{(1-x)(1+x)}=\lim_{h\to 1} \frac{-\sqrt{2-x}}{(1+x)}=\frac{-1}{2}$$


 

4.- $\displaystyle \lim_{h\to 1} \frac{1}{1-x}-\frac{3}{1-x^3}$

$$\displaystyle \lim_{h\to 1} \frac{1}{1-x}-\frac{3}{1-x^3}=\frac{1}{1-1}-\frac{3}{1-1^3}=\frac{1}{0}-\frac{3}{0}= \infty-\infty$$

Obviamente un indeterminado, pero

$$\displaystyle \lim_{h\to 1} \frac{1}{1-x}-\frac{3}{1-x^3}=\lim_{h\to 1} \frac{1}{1-x}-\frac{3}{(1-x)(1+x+x^2)}\\=\lim_{h\to 1} \frac{1+x+x^2}{(1-x)(1+x+x^2)}-\frac{3}{(1-x)(1+x+x^2)} \\= \lim_{h\to 1} \frac{1+x+x^2-3}{(1-x)(1+x+x^2)} \\= \lim_{h\to 1} \frac{x^2+x-2}{(1-x)(1+x+x^2)}\\= \lim_{h\to 1} \frac{(x-1)(x+2)}{(1-x)(1+x+x^2)}\\= \lim_{h\to 1} \frac{-(1-x)(x+2)}{(1-x)(1+x+x^2)}\\= \lim_{h\to 1} \frac{-(x+2)}{1+x+x^2}=\frac{-3}{3}=-1$$


 

5.- $\displaystyle \lim_{h\to 1}\frac{x+2}{x^2-5x+4}+\frac{x-4}{3(x^2-3x+2)}$

$$\displaystyle \lim_{h\to 1}\frac{x+2}{x^2-5x+4}+\frac{x-4}{3(x^2-3x+2)}=\frac{1+2}{1-5+4}+\frac{1-4}{3(1-3+2)}=\frac{1}{0}-\frac{1}{0}= \infty-\infty$$

también indeterminado, pero

$$\displaystyle \lim_{h\to 1}\frac{x+2}{x^2-5x+4}+\frac{x-4}{3(x^2-3x+2)}=\lim_{h\to 1}\frac{x+2}{(x-4)(x-1)}+\frac{x-4}{3(x-2)(x-1)}\\=\lim_{h\to 1}\frac{3(x+2)(x-2)+(x-4)(x-4)}{3(x-4)(x-2)(x-1)}\\= \lim_{h\to 1}\frac{3(x^2-4)+(x-4)^2}{3(x-4)(x-2)(x-1)}\\= \lim_{h\to 1}\frac{3(x^2-4)+(x-4)^2}{3(x-4)(x-2)(x-1)}\\= \lim_{h\to 1}\frac{3x^2-12+x^2-8x+16}{3(x-4)(x-2)(x-1)}\\=\lim_{h\to 1}\frac{4x^2-8x+4}{3(x-4)(x-2)(x-1)}\\= \lim_{h\to 1}\frac{4(x^2+1)}{3(x-4)(x-2)(x-1)}\\=\lim_{h\to 1}\frac{4(x-1)^2}{3(x-4)(x-2)(x-1)}\\=\lim_{h\to 1}\frac{4(x-1)}{3(x-4)(x-2)}=\frac{0}{24}=0$$


En desarrollo. 

6.- $\displaystyle \lim_{h\to \infty}{2+x-3x^3}{5-x^2+3x^3}$$

7.- $\displaystyle \lim_{h\to \infty}\frac {x^3}{x^2+1}$

8.- $\displaystyle \lim_{h\to \infty}\frac {\sqrt{x^2+1}-3\sqrt{x}}{\sqrt[4]{x^3+5x-x}-x}$

9.- $\displaystyle \lim_{h\to 0}\frac {\sqrt{1+x^2}-1}{x}$

10.-$\displaystyle \lim_{h\to 5}\frac {\sqrt{x-1}-2}{x-5}$

11.- $\displaystyle \lim_{h\to 1}\frac {\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}$

12.-  $\displaystyle \lim_{h\to \infty} \sqrt{(x+m)(x+n}-x$

13.- $\displaystyle \lim_{h\to 2}\frac{1}{2-x}-\frac{3}{8-x^3}$

14.- $\displaystyle \lim_{h\to \frac {1}{2}}\frac {8x^2-1}{6x^2-5x+1}$

15.- $\displaystyle \lim_{h\to \infty}\frac {3x^4-2x+1}{3x^2+6x-2}$

16.- $\displaystyle \lim_{h\to 0}\frac {\sqrt{x^2+4}-2}{\sqrt{x^2+9}-3}$

17.- $\displaystyle \lim_{h\to 4}\frac {\sqrt{1+2x}-3}{x-4}$

18.- $\displaystyle \lim_{h\to \infty}\frac{3^x+x}{3^x-2x}$

19.-  $\displaystyle \lim_{h\to \infty}\frac {\sqrt[3]{x^2+1}}{x-2}$

20.- $\displaystyle \lim_{h\to \infty}\frac{2x^2+1}{\sqrt[3]{x^7+2x}}$

21.- $\displaystyle \lim_{h\to 0}\frac {\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$

22.- $\displaystyle \lim_{h\to 1}\frac{1-\sqrt[3]{x}}{1-\sqrt{x}}$

 

 

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