Clase 11 RD Sharma Solutions – Capítulo 29 Límites – Ejercicio 29.8

Pregunta 20. lím _x→1 [(1 + cosπx)/(1 – x) ² ]

Solución:

Tenemos,

lím _x→1 [(1 + cosπx)/(1 – x) ² ]

Aquí,

= Lim _h→0 [{1 + cosπ(1 + h)}/{1 – (1 + h)} ² ]

= Lim _h→0 [(1 – cosπh)/h2]

= Lím _h→0 [2sen ² (πh/2)/h ² ]

= $2\lim_{h\to0}[\frac{sin^2(\frac{πh}{2})}{(\frac{2}{π})^2×(\frac{πh}{2})^2}]$

= 2π ² /4

= π ² /2

Pregunta 21. lím _x→1 [(1 – x ² )/senπx]

Solución:

Tenemos,

límite _x→1 [(1 – x ² )/senπx]

Aquí,

= lím _h→0 [{1 – (1 – h) ² }/senπ(1 – h)]

= lím _h→0 [(2h – h ² )/-senπh]

= -lím _h→0 [{h(2 – h)}/senπh]

= $\lim_{h\to0}[\frac{(2-h)}{\frac{sinπh}{h}}]$

= $\lim_{h\to0}[\frac{(2-h)}{(\frac{sinπh}{πh})×π}]$

= (2 – 0)/π

= 2/π

Pregunta 22. lím _x→π/4 [(1 – sen2x)/(1 + cos4x)]

Solución:

Tenemos,

lím _x→π/4 [(1 – sen2x)/(1 + cos4x)]

Aquí, π/4

= lím _h→0 [{1 – sin2(π/4 – h)}/{1 + cos4(π/4 – h)}]

= lím _h→0 [{1 – sin(π/2 – 2h)}/{1 + cos(π – 4h)}]

= lím _h→0 [(1 – cos2h)/(1 – cos4h)]

= lím _h→0 [2sen ² h/2sen ² 2h]

= $\lim_{h\to0}[\frac{\frac{sin^2h}{h^2}}{\frac{sin^22h}{4h^2}×4}]$

= (1/4)

Pregunta 23. lim _x→π [(1 + cosx)/tan ² x]

Solución:

Tenemos,

lím _x→π [(1 + cosx)/tan ² x]

Aquí, π

= lím _h→0 [{1+cos(π + h)}/tan ² (π + h)]

= lim _h→0 [(1 – cosh)/tan ² h]

= lím _h→0 [{2sen ² (h/2)}/tan ² h]

= $2×\lim_{h\to0}[\frac{sin^2(\frac{h}{2})}{(\frac{h}{2})^2×(\frac{2}{h})^2}]×\lim_{h\to0}[\frac{1}{\frac{tan^2h}{h^2}×h^2}]$

= 2/4

= 1/2

Pregunta 24. lim _n→∞ [nsen(π/4n)cos(π/4n)]

Solución:

Tenemos,

lím _n→∞ [nsen(π/4n)cos(π/4n)]

= lím _n→∞ [nsen(π/4n)]Limn→∞[cos(π/4n)]

= $\lim_{n\to∞}[\frac{nsin\frac{π}{4n}}{(\frac{π}{4n})}]×(\frac{π}{4n})×1$

= $(\frac{π}{4})\lim_{n\to∞}[\frac{sin\frac{π}{4n}}{(\frac{π}{4n})}]$

Sea, y = (π/4n)

Si n→∞, y→0.

= (π/4).Lim _y→0 [seno/y]

= (π/4)

Pregunta 25. lim _n→∞ [2 ^n-1 sin(a/2 ⁿ )]

Solución:

Tenemos,

lím _n→∞ [2 ^n-1 sin(a/2 ⁿ )]

= $\lim_{n\to∞}[\frac{2^n}{2}×sin(\frac{a}{2^n})]$

= $\lim_{n\to∞}[\frac{2^n}{2}×\frac{sin(\frac{a}{2^n})}{\frac{a}{2^n}}×\frac{a}{2^n}]$

= $(\frac{a}{2})\lim_{n\to∞}[\frac{sin\frac{a}{2^n}}{(\frac{a}{2^n})}]$

Sea, y = (a/2 ⁿ )

Si n→∞, y→0

= (a/2).Lim _y→0 [seno/y]

= (a/2)

Pregunta 26. lim _n→∞ [sen(a/2 ⁿ )/sen(b/2 ⁿ )]

Solución:

Tenemos,

lím _n→∞ [sen(a/2 ⁿ )/sen(b/2 ⁿ )]

= $\lim_{n\to∞}[\frac{sin(\frac{a}{2^n})}{(\frac{a}{2^n})}×(\frac{a}{2^n})]\lim_{n\to∞}[\frac{}{\frac{sin(\frac{b}{2^n})}{(\frac{b}{2^n})}×(\frac{b}{2^n})}]$

Sea, y = (a/2 ⁿ ) y z = (b/2 ⁿ )

Si n→∞, y→0 y z→0

= $\frac{y}{z}\lim_{y\to0}[\frac{siny}{y}]\lim_{z\to0}[\frac{y}{siny}]$

= $\frac{\frac{a}{2^n}}{\frac{b}{2^n}}$

= (a/b)

Pregunta 27. lím _x→-1 [(x ² – x – 2)/{(x ² + x) + sin(x + 1)}]

Solución:

Tenemos,

límite _x→-1 [(x ² – x – 2)/{(x ² + x) + sin(x + 1)}]

= límite _x→-1 [(x ² – x – 2)/{x(x + 1) + sin(x + 1)}]

= límite _x→-1 [(x – 2)(x + 1)/{x(x + 1) + sin(x + 1)}]

Sea, y = x + 1

Si x→-1, entonces y→0

= lím _y→0 [y(y – 3)/{y(y – 1) + seno}]

= $\lim_{y\to0}[\frac{(y-3)}{(y-1)+\frac{siny}{y}}]$

= (0 – 3)/{(0 – 1) + 1}

= -3/0

= ∞

Pregunta 28. lím _x→2 [(x ² – x – 2)/{(x ² – 2x) + sin(x – 2)}]

Solución:

Tenemos,

límite _x→2 [(x ² – x – 2)/{(x ² – 2x) + sin(x – 2)}]

= límite _x→2 [{(x – 2)(x + 1)}/{x(x + 1) + sin(x + 1)}]

Sea, y = x – 2

Si x→2, entonces y→0

= lím _y→0 [y(y + 3)/{y(y + 2) + seno}]

= $\lim_{y\to0}[\frac{(y+3)}{(y+2)+\frac{siny}{y}}]$

= (0 + 3)/{(0 + 1) + 1}

= 3/3

= 1

Pregunta 29. lím _x→1 [(1 – x)tan(πx/2)]

Solución:

Tenemos,

límite _x→1 [(1 – x)tan(πx/2)]

Aquí,

= lím _h→0 [{1 – (1 – h)}tan{π/2(1 – h)}]

= lím _h→0 [htan{π/2-πh/2)}

= lím _h→0 [hcot(πh/2)]

= $\lim_{h\to0}[\frac{h}{tan(\frac{πh}{2})}]$

= $\lim_{h\to0}[\frac{1}{\frac{tan(\frac{πh}{2})}{\frac{πh}{2}}×\frac{π}{2}}]$

= $\frac{1}{\frac{π}{2}}$

= (2/π)

Pregunta 30. lím _x→π/4 [(1 – tanx)/(1 – √2senx)]

Solución:

Tenemos,

lím _x→π/4 [(1 – tanx)/(1 – √2senx)]

Sobre la racionalización del denominador.

= lím _x→π/4 [{(1 – tanx)(1 – √2senx)}/(1 – 2sen ² x)]

= $\lim_{x\to \frac{π}{4}}[\frac{(1-\frac{sinx}{cosx})(1+\sqrt2sinx)}{cos2x}]$

= $\lim_{x\to \frac{π}{4}}[\frac{(cosx-sinx)(1+\sqrt2sinx)}{cosx.cos2x}]$

= $\lim_{x\to \frac{π}{4}}[\frac{(cosx-sinx)(1+\sqrt2sinx)}{cosx.(cos^2x-sin^2x)}]$

= $\lim_{x\to \frac{π}{4}}[\frac{(cosx-sinx)(1+\sqrt2sinx)}{cosx.(cosx-sinx)(cosx+sinx)}]$

= $\frac{1+\sqrt2×\frac{1}{\sqrt2}}{(\frac{1}{\sqrt2})(\frac{1}{\sqrt2}+\frac{1}{\sqrt2})}$

= 2/1

= 2

Pregunta 31. lim _x→π [{√(2 + cosx) – 1}/(π – x) ² ]

Solución:

Tenemos,

límite _x→π [{√(2 + cosx) – 1}/(π – x) ² ]

Sea, y = [π – x]

Aquí, x→π, y→0

= $\lim_{y\to0}[\frac{\sqrt{2+cos(π-y)}-1}{y^2}]$

= lim _y→0 [{√(2 – acogedor) – 1}/y2]

Al racionalizar el numerador, obtenemos

= $\lim_{y\to0}[\frac{2-cosy-1}{y^2(\sqrt{2-cosy}-1)}]$

= lim _y→0 [{1 – acogedor}/y2{√(2 – acogedor) – 1}]

= $\lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{y^2(\sqrt{2-cosx}+1)}]$

= $\lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{(\frac{y}{2})^2(\sqrt{2-cosx}+1)×4}]$

= 2 × (1/4) × {1/(1 + 1)}

= (1/4)

Pregunta 32. lím _x→π/4 [(√cosx – √senx)/(x – π/4)]

Solución:

Tenemos,

lím _x→π/4 [(√cosx – √senx)/(x – π/4)]

Al racionalizar el numerador, obtenemos

= lím _x→π/4 [(cosx – senx)/{(√cosx + √senx)(x – π/4)}]

= $\lim_{h\to0}[\frac{cos(\frac{π}{4}+h)-sin(\frac{π}{4}+h)}{(\frac{π}{4}+h-\frac{π}{4})(\sqrt{cos(\frac{π}{4}+h)}+\sqrt{sin(\frac{π}{4}+h)}}]$

= $\lim_{h\to0}[\frac{-2×\frac{1}{\sqrt2}×sinh}{h(\sqrt{cos(\frac{π}{4}+h)}+\sqrt{sin(\frac{π}{4}+h)})}]$

= $\lim_{h\to0}[\frac{-\sqrt2×sinh}{h(\sqrt{cos(\frac{π}{4}+h)}+\sqrt{sin(\frac{π}{4}+h)})}]$

= $-\sqrt2[\frac{1}{(\frac{1}{\sqrt2})^{\frac{1}{2}}+(\frac{1}{\sqrt2})^{\frac{1}{2}}}]$

Pregunta 33. lím _x→1 [(1 – 1/x)/sinπ(x – 1)]

Solución:

Tenemos,

límite _x→1 [(1 – 1/x)/senπ(x – 1)]

= límite _x→1 [(x – 1)/x{sinπ(x – 1)}]

Sea, y = x – 1

Si x→1, entonces y→0

= lím _y→0 [y/{(y + 1)sen(πy)}]

= $\lim_{y\to0}[\frac{1}{\frac{(y+1).sin(πy)}{y}}]$

= $\lim_{y\to0}[\frac{1}{\frac{(y+1).sin(πy)}{πy}×π}]$

= 1/{(1 + 0) × 1 × π}

= 1/π

Pregunta 34. lim _x→π/6 [(cot ² x – 3)/(cosecx – 2)]

Solución:

Tenemos,

lím _x→π/6 [(cot ² x – 3)/(cosecx – 2)]

= lím _x→π/6 [(cosec ² x – 1 – 3)/(cosecx – 2)]

= lím _x→π/6 [(cosec ² x – 2 ² )/(cosecx – 2)]

= lím _x→π/6 [{(cosecx + 2)(cosecx – 2)}/(cosecx – 2)]

= lím _x→π/6 [(cosecx + 2)]

= cosec(π/6) + 2

= 2 + 2

= 4

Pregunta 35. lím _x→π/4 [(√2 – cosx – senx)/(4x – π) ² ]

Solución:

Tenemos,

lím _x→π/4 [(√2 – cosx – senx)/(4x – π) ² ]

= lím _x→π/4 [(√2 – cosx – senx)/{4 ² (π/4 – x) ² }]

= $\lim_{x\to \frac{π}{2}}\frac{\sqrt2[1-(\frac{1}{\sqrt2}.cosx+\frac{1}{\sqrt2}.sinx)]}{4^2(\frac{π}{4}-x)^2}$

= $\lim_{x\to \frac{π}{2}}\frac{\sqrt{2}[1-cos(\frac{π}{4}-x)]}{4^2(\frac{π}{4}-x)^2}$

= $\lim_{x\to \frac{π}{2}}\frac{[2\sqrt{2}sin^2\frac{(\frac{π}{4}-x)}{2}]}{4^2(\frac{π}{4}-x)^2}$

= $2\sqrt{2}\lim_{x\to \frac{π}{2}}\frac{[sin^2\frac{(\frac{π}{4}-x)}{2}]}{4^2×\frac{(\frac{π}{4}-x)}{4}^2×4}$

= 2√2/4 ³

= (2√2 × √2)/(4 ³ √2)

= 4/(4 ³ √2)

= 1/(16√2)

Pregunta 36. lim _x→π/2 [{(π/2 – x)senx – 2cosx}/{(π/2 – x) + cotx}]

Solución:

Tenemos,

lím _x→π/2 [{(π/2 – x)senx – 2cosx}/{(π/2 – x) + cotx}]

= $\lim_{h\to0}\frac{[\frac{π}{2}-(\frac{π}{2}-h)]sin(\frac{π}{2}-h)-2cos(\frac{π}{2}-h)}{[\frac{π}{2}-(\frac{π}{2}-h)]+cot(\frac{π}{2}-h)}$

= lím _h→0 [(hcosh-2sinh)/(h+tanh)]

= $\lim_{h\to0}[\frac{cosh-2\frac{sinh}{h}}{1+\frac{tanh}{h}}]$ (Al dividir el numerador y el denominador por h)

= (1 – 2)/(1 + 1)

= -1/2

Pregunta 37. lim _x→π/4 [(cosx – senx)/{(π/4 – x)(cosx + senx)}]

Solución:

Tenemos,

lím _x→π/4 [(cosx – senx)/{(π/4 – x)(cosx + senx)}]

Al dividir el numerador y el denominador por √2, obtenemos

= $\lim_{x\to \frac{π}{4}}[\frac{\frac{cosx}{\sqrt2}-\frac{sinx}{\sqrt2}}{(\frac{π}{4}-x)(\frac{cosx+sinx}{\sqrt2})}]$

= $\lim_{x\to \frac{π}{4}}[\frac{{\sqrt2}(sin\frac{π}{4}.cosx-cos\frac{π}{4}.sinx)}{(\frac{π}{4}-x)(cosx+sinx)}]$

= $\lim_{x\to \frac{π}{4}}[\frac{{\sqrt2}[sin(\frac{π}{4}-x)]}{(\frac{π}{4}-x)(cosx+sinx)}]$

= $\sqrt2\lim_{x\to \frac{π}{4}}[\frac{[sin(\frac{π}{4}-x)]}{(\frac{π}{4}-x)}]\lim_{x\to \frac{π}{4}}[\frac{1}{(cosx+sinx)}]$

= $\frac{\sqrt2}{\frac{1}{\sqrt2}+\frac{1}{\sqrt2}}$

= (√2 × √2)/2

= 1

Pregunta 38. lim _x→π [{1 – sen(x/2)}/{cos(x/2)(cosx/4 – senx/4}]

Solución:

Tenemos,

lím _x→π [{1 – sen(x/2)}/{cos(x/2)(cosx/4 – senx/4}]

Sea, x = π + h

Si x→π, entonces h→0

= $\lim_{h\to0}[\frac{1-sin(\frac{π+h}{2})}{cos(\frac{π+h}{2})[cos(\frac{π+h}{4})-sin(\frac{π+h}{4})]}]$

= $\lim_{h\to0}[\frac{1-sin(\frac{π}{2}+\frac{h}{2})}{cos(\frac{π}{4}+\frac{h}{2})[cos(\frac{π}{4}+\frac{h}{4})-sin(\frac{π}{4}+\frac{h}{4})]}]$

= $\lim_{h\to0}[\frac{1-cos(\frac{h}{2})}{-sin(\frac{h}{2})[-\sqrt{2}sin(\frac{h}{2})]}]$

= $\lim_{h\to0}[\frac{2sin^2(\frac{h}{4})}{-sin(\frac{h}{2})[-\sqrt{2}sin(\frac{h}{4})]}]$

= $\lim_{h\to0}[\frac{2sin^2(\frac{h}{4})}{2sin(\frac{h}{4})cos(\frac{h}{4})[\sqrt{2}sin(\frac{h}{4})]}]$

= $(\frac{1}{\sqrt2})\lim_{h\to0}\frac{1}{cos(\frac{h}{4})}$

= 1/√2

Publicación traducida automáticamente

Artículo escrito por vivekray59 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

Clase 11 RD Sharma Solutions – Capítulo 29 Límites – Ejercicio 29.8 | conjunto 2

Pregunta 20. lím _x→1 [(1 + cosπx)/(1 – x) ² ]

Pregunta 21. lím _x→1 [(1 – x ² )/senπx]

Pregunta 22. lím _x→π/4 [(1 – sen2x)/(1 + cos4x)]

Pregunta 23. lim _x→π [(1 + cosx)/tan ² x]

Pregunta 24. lim _n→∞ [nsen(π/4n)cos(π/4n)]

Pregunta 25. lim _n→∞ [2 ^n-1 sin(a/2 ⁿ )]

Pregunta 26. lim _n→∞ [sen(a/2 ⁿ )/sen(b/2 ⁿ )]

Pregunta 27. lím _x→-1 [(x ² – x – 2)/{(x ² + x) + sin(x + 1)}]

Pregunta 28. lím _x→2 [(x ² – x – 2)/{(x ² – 2x) + sin(x – 2)}]

Pregunta 29. lím _x→1 [(1 – x)tan(πx/2)]

Pregunta 30. lím _x→π/4 [(1 – tanx)/(1 – √2senx)]

Pregunta 31. lim _x→π [{√(2 + cosx) – 1}/(π – x) ² ]

Pregunta 32. lím _x→π/4 [(√cosx – √senx)/(x – π/4)]

Pregunta 33. lím _x→1 [(1 – 1/x)/sinπ(x – 1)]

Pregunta 34. lim _x→π/6 [(cot ² x – 3)/(cosecx – 2)]

Pregunta 35. lím _x→π/4 [(√2 – cosx – senx)/(4x – π) ² ]

Pregunta 36. lim _x→π/2 [{(π/2 – x)senx – 2cosx}/{(π/2 – x) + cotx}]

Pregunta 37. lim _x→π/4 [(cosx – senx)/{(π/4 – x)(cosx + senx)}]

Pregunta 38. lim _x→π [{1 – sen(x/2)}/{cos(x/2)(cosx/4 – senx/4}]

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Pregunta 20. lím x→1 [(1 + cosπx)/(1 – x) 2 ]

Pregunta 21. lím x→1 [(1 – x 2 )/senπx]

Pregunta 22. lím x→π/4 [(1 – sen2x)/(1 + cos4x)]

Pregunta 23. lim x→π [(1 + cosx)/tan 2 x]

Pregunta 24. lim n→∞ [nsen(π/4n)cos(π/4n)]

Pregunta 25. lim n→∞ [2 n-1 sin(a/2 n )]

Pregunta 26. lim n→∞ [sen(a/2 n )/sen(b/2 n )]

Pregunta 27. lím x→-1 [(x 2 – x – 2)/{(x 2 + x) + sin(x + 1)}]

Pregunta 28. lím x→2 [(x 2 – x – 2)/{(x 2 – 2x) + sin(x – 2)}]

Pregunta 29. lím x→1 [(1 – x)tan(πx/2)]

Pregunta 30. lím x→π/4 [(1 – tanx)/(1 – √2senx)]

Pregunta 31. lim x→π [{√(2 + cosx) – 1}/(π – x) 2 ]

Pregunta 32. lím x→π/4 [(√cosx – √senx)/(x – π/4)]

Pregunta 33. lím x→1 [(1 – 1/x)/sinπ(x – 1)]

Pregunta 34. lim x→π/6 [(cot 2 x – 3)/(cosecx – 2)]

Pregunta 35. lím x→π/4 [(√2 – cosx – senx)/(4x – π) 2 ]

Pregunta 36. lim x→π/2 [{(π/2 – x)senx – 2cosx}/{(π/2 – x) + cotx}]

Pregunta 37. lim x→π/4 [(cosx – senx)/{(π/4 – x)(cosx + senx)}]

Pregunta 38. lim x→π [{1 – sen(x/2)}/{cos(x/2)(cosx/4 – senx/4}]

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Pregunta 20. lím _x→1 [(1 + cosπx)/(1 – x) ² ]

Pregunta 21. lím _x→1 [(1 – x ² )/senπx]

Pregunta 22. lím _x→π/4 [(1 – sen2x)/(1 + cos4x)]

Pregunta 23. lim _x→π [(1 + cosx)/tan ² x]

Pregunta 24. lim _n→∞ [nsen(π/4n)cos(π/4n)]

Pregunta 25. lim _n→∞ [2 ^n-1 sin(a/2 ⁿ )]

Pregunta 26. lim _n→∞ [sen(a/2 ⁿ )/sen(b/2 ⁿ )]

Pregunta 27. lím _x→-1 [(x ² – x – 2)/{(x ² + x) + sin(x + 1)}]

Pregunta 28. lím _x→2 [(x ² – x – 2)/{(x ² – 2x) + sin(x – 2)}]

Pregunta 29. lím _x→1 [(1 – x)tan(πx/2)]

Pregunta 30. lím _x→π/4 [(1 – tanx)/(1 – √2senx)]

Pregunta 31. lim _x→π [{√(2 + cosx) – 1}/(π – x) ² ]

Pregunta 32. lím _x→π/4 [(√cosx – √senx)/(x – π/4)]

Pregunta 33. lím _x→1 [(1 – 1/x)/sinπ(x – 1)]

Pregunta 34. lim _x→π/6 [(cot ² x – 3)/(cosecx – 2)]

Pregunta 35. lím _x→π/4 [(√2 – cosx – senx)/(4x – π) ² ]

Pregunta 36. lim _x→π/2 [{(π/2 – x)senx – 2cosx}/{(π/2 – x) + cotx}]

Pregunta 37. lim _x→π/4 [(cosx – senx)/{(π/4 – x)(cosx + senx)}]

Pregunta 38. lim _x→π [{1 – sen(x/2)}/{cos(x/2)(cosx/4 – senx/4}]