Clase 11 RD Sharma Solutions – Capítulo 29 Límites – Ejercicio 29.3

Posted on by Rudeus Greyrat

Pregunta 1. Evaluar \displaystyle\lim_{x\to-5}\frac{2x^2+9x-5}{x+5}

Solución:

\displaystyle\lim_{x\to-5}\frac{2x^2+9x-5}{x+5}\\ =\displaystyle\lim_{x\to-5}\frac{(x+5)(2x-1)}{(x+5)}\\

=\displaystyle\lim_{x\to-5}{(2x-1)}  = 2(-5) – 1

= -10 – 1 

= -11

Pregunta 2. Evaluar \displaystyle\lim_{x\to3}\frac{x^2-4x+3}{x^2-2x-3}

Solución:

\displaystyle\lim_{x\to3}\frac{x^2-4x+3}{x^2-2x-3}\\ =\displaystyle\lim_{x\to3}\frac{x^2-3x-x+3}{x^2+x-3x-3}\\ =\displaystyle\lim_{x\to3}\frac{x(x-1)-3(x-1)}{x(x+1)-3(x+1)}\\ =\displaystyle\lim_{x\to3}\frac{(x-1)(x-3)}{(x+1)(x-3)}\\ =\displaystyle\lim_{x\to3}\frac{x-1}{x+1}\\ =\frac{3-1}{3+1}\\ =\frac{2}{4}\\ =\frac{1}{2}

Pregunta 3. Evaluar \displaystyle\lim_{x\to3}\frac{x^4-81}{x^2-9}

Solución:

\displaystyle\lim_{x\to3}\frac{x^4-81}{x^2-9}\\ =\displaystyle\lim_{x\to3}\frac{(x^2-9)(x^2+9)}{(x^2-9)}\\ =\displaystyle\lim_{x\to3}x^2+9\\

= (3) 2 + 9 = 9 + 9 = 18

Pregunta 4. Evaluar \displaystyle\lim_{x\to2}\frac{x^3-8}{x^2-4}

Solución:

\displaystyle\lim_{x\to2}\frac{x^3-8}{x^2-4}\\ =\displaystyle\lim_{x\to2}\frac{(x-2)(x^2+4+2x)}{(x-2)(x+2)}\\ =\frac{(2)^2+4+2(2)}{2+2}\\ =\frac{4+4+4}{4}\\ =\frac{12}{4}

= 3

Pregunta 5. Evaluar \displaystyle\lim_{x\to-\frac{1}{2}}\frac{8x^3+1}{2x+1}

Solución:

\displaystyle\lim_{x\to-\frac{1}{2}}\frac{8x^3+1}{2x+1}\\ =\displaystyle\lim_{x\to-\frac{1}{2}}\frac{8\left(x^3+\frac{1}{8}\right)}{2\left(x+\frac{1}{2}\right)}\\ =\frac{8}{2}\displaystyle\lim_{x\to-\frac{1}{2}}\frac{\left(x^3+\left(\frac{1}{2}\right)^3\right)}{x+\frac{1}{2}}\\ =4\displaystyle\lim_{x\to-\frac{1}{2}}\frac{\left(x+\frac{1}{2}\right)\left(x^2+\frac{1}{4}-\frac{1}{2}x\right)}{\left(x+\frac{1}{2}\right)}\\ =4\left(\left(\frac{-1}{2}\right)^2+\frac{1}{4}-\frac{1}{2}\left(\frac{1}{2}\right)\right)\\ =4\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)

= 3

Pregunta 6. Evaluar \displaystyle\lim_{x\to4}\frac{x^2-7x+12}{x^2-3x-4}

Solución:

\displaystyle\lim_{x\to4}\frac{x^2-7x+12}{x^2-3x-4}\\ =\displaystyle\lim_{x\to4}\frac{x^2-3x-4x+12}{x^2+x-4x-4}\\ =\displaystyle\lim_{x\to4}\frac{x(x-3)-4(x-3)}{x(x+1)-1(x+1)}\\ =\displaystyle\lim_{x\to4}\frac{(x-3)(x-4)}{(x-4)(x+1)}\\ =\displaystyle\lim_{x\to4}\frac{x-3}{x+1}\\ =\frac{4-3}{4+1}\\ =\frac{1}{5}

Pregunta 7. Evaluar \displaystyle\lim_{x\to2}\frac{x^4-16}{x-2}

Solución:

\displaystyle\lim_{x\to2}\frac{x^2-16}{x-2}\\ =\displaystyle\lim_{x\to2}\frac{(x^2-4)(x^2+4)}{(x-2)}\\ =\displaystyle\lim_{x\to2}\frac{(x-2)(x+2)(x^2+4)}{(x-2)}\\ =\displaystyle\lim_{x\to2}(x+2)(x^2+4)

= (2 + 2)(4 + 4)

= 4(8)

= 32

Pregunta 8. Evaluar \displaystyle\lim_{x\to5}\frac{x^2-9x+20}{x^2-6x+5}

Solución:

\displaystyle\lim_{x\to5}\frac{x^2-9x+20}{x^2-6x+5}\\ =\displaystyle\lim_{x\to5}\frac{x^2-4x-5x+20}{x^2-x-5x+5}\\ =\displaystyle\lim_{x\to5}\frac{x(x-4)-5(x-4)}{x(x-1)-5(x-1)}\\ =\displaystyle\lim_{x\to5}\frac{(x-5)(x-4)}{(x-5)(x-1)}\\ =\displaystyle\lim_{x\to5}\frac{x-4}{x-1}\\ =\frac{5-4}{5-1}\\ =\frac{1}{4}

Pregunta 9. \displaystyle\lim_{x\to-1}\frac{x^3+1}{x+1}

Solución:

\displaystyle\lim_{x\to-1}\frac{x^3+1}{x+1}\\ =\displaystyle\lim_{x\to-1}\frac{(x+1)(x^2-x+1)}{(x+1)}\ \ \ \ \ \ [a^3+b^3=(a+b)(a^2+b^2-ab)]\\ =\displaystyle\lim_{x\to-1}(x^2-x+1)

= (-1) 2 – (-1) + 1

= 1 + 1 + 1

= 3

Pregunta 10. Evaluar \displaystyle\lim_{x\to5}\frac{x^3-125}{x^2-7x+10}

Solución: 

\displaystyle\lim_{x\to5}\frac{x^3-125}{x^2-7x+10}\\ =\displaystyle\lim_{x\to5}\frac{(x-5)(x^2+25+5x)}{(x-2)(x-5)}\\ =\frac{(5)^2+25+5(5)}{(5-2)}\\ =\frac{25+25+25}{3}\\ =\frac{75}{3}

Pregunta 11. Evaluar \displaystyle\lim_{x\to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4}

Solución:

\displaystyle\lim_{x\to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4}

=\displaystyle\lim_{x\to\sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{x^2+2\sqrt{2}x-\sqrt{2}x-4}\\ =\displaystyle\lim_{x\to\sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{x(x+2\sqrt{2}-\sqrt{2(x+2\sqrt{2}})}\\ =\displaystyle\lim_{x\to\sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{(x+2\sqrt{2})(x-\sqrt{2})}\\ =\frac{\sqrt{2}+\sqrt{2}}{\sqrt{2}+2\sqrt{2}}\\ =\frac{2\sqrt{2}}{3\sqrt{2}}\\ =\frac{2}{3}

Pregunta 12. Evaluar \displaystyle\lim_{x\to\sqrt{3}}\frac{x^2-3}{x^2+3\sqrt{3}-12}

Solución:

\displaystyle\lim_{x\to\sqrt{3}}\frac{x^2-3}{x^2+3\sqrt{3}-12}

=\displaystyle\lim_{x\to\sqrt{3}}\frac{(x-\sqrt{3})(x+\sqrt{3})}{x^2+4\sqrt{3x}-\sqrt{3x}-12}\\ =\displaystyle\lim_{x\to\sqrt{3}}\frac{(x-\sqrt{3})(x+\sqrt{3})}{x(x+4\sqrt{3})-\sqrt{3}(x+4\sqrt{3})}\\ =\displaystyle\lim_{x\to\sqrt{3}}\frac{(x-\sqrt{3})(x+\sqrt{3})}{(x-\sqrt{3})(x+4\sqrt{3})}\\ =\frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}+4\sqrt{3}}\\ =\frac{2\sqrt{3}}{5\sqrt{3}}\\ =\frac{2}{5}

Pregunta 13. Evaluar \displaystyle\lim_{x\to\sqrt{3}}\frac{x^4-9}{x^2+4\sqrt{3}x-15}

Solución:

\displaystyle\lim_{x\to\sqrt{3}}\frac{x^4-9}{x^2+4\sqrt{3}x-15}

=\displaystyle\lim_{x\to\sqrt{3}}\frac{(x-\sqrt{3})(x+\sqrt{3})(x^2+3)}{(x-\sqrt{3})(x+5\sqrt{3})}\\ =\displaystyle\lim_{x\to\sqrt{3}}\frac{(x+\sqrt{3})(x^2+3)}{(x+5\sqrt{3})}\\ =\frac{(\sqrt{3}+\sqrt{3})(3+3)}{(\sqrt{3}+5\sqrt{3})}\\ =\frac{(2\sqrt{3})(6)}{6\sqrt{3}}

= 2

Pregunta 14. Evaluar \displaystyle\lim_{x\to\sqrt{2}}{\left(\frac{x}{x-2}-\frac{4}{x^2-2x}\right)}

Solución:

\displaystyle\lim_{x\to\sqrt{2}}{\left(\frac{x}{x-2}-\frac{4}{x^2-2x}\right)}

=\displaystyle\lim_{x\to2}{\left(\frac{x}{x-2}-\frac{4}{x(x-2)}\right)}\\ =\displaystyle\lim_{x\to2}{\left(\frac{x(x)-4}{x(x-2)}\right)}\\ =\displaystyle\lim_{x\to2}{\left(\frac{x^2-4}{x(x-2)}\right)}\\ =\displaystyle\lim_{x\to2}{\left(\frac{(x-2)(x+2)}{x(x-2)}\right)}\\ =\displaystyle\lim_{x\to2}\frac{(x+2)}{x}\\ =\frac{2+2}{2}\\ =\frac{4}{2}

= 2

Publicación traducida automáticamente

Artículo escrito por yashchuahan y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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