Pregunta 1. lím x→π/2 [π/2 – x].tanx
Solución:
Tenemos,
lím x→π/2 [π/2 – x].tanx
Consideremos, y = [π/2 – x]
Aquí, x→π/2, y→0
= lim y→0 [y.tan(π/2 – y)]
= lím y→0 [y.{sen(π/2 – y)/cos(π/2 – y)]
= lim y→0 [y.{cosy/seny}]
= lím y→0 [y/seno].cosy
= lim y→0 [cosy] [Ya que, lim y→0 [seny/y] = 1]
= 1
Pregunta 2. lím x→π/2 [sen2x/cosx]
Solución:
Tenemos,
lím x→π/2 [sen2x/cosx]
= lím x→π/2 [2senx.cosx/cosx]
= 2Lím x→π/2 [senx]
= 2
Pregunta 3. lím x→π/2 [cos 2 x/(1 – senx)]
Solución:
Tenemos,
lím x→π/2 [cos 2 x/(1 – senx)]
= lím x→π/2 [(1 – sen 2 x)/(1 – senx)]
= lím x→π/2 [(1 – senx)(1 + senx)/(1 – senx)]
= lím x→π/2 [(1 + senx)]
= 1 + 1
= 2
Pregunta 4. lím x→π/2 [(1 – senx)/cos 2 x]
Solución:
Tenemos,
lím x→π/2 [(1 – senx)/cos 2 x]
= lím x→π/2 [(1 – senx)/(1 – sen 2 x)]
= lím x→π/2 [(1 – senx)/(1 – senx)(1 + senx)]
= lím x→π/2 [1/(1 + senx)]
= 1/(1 + 1)
= 1/2
Pregunta 5. lim x→a [(cosx – cosa)/(x – a)]
Solución:
Tenemos,
lím x→a [(cosx – cosa)/(x – a)]
=
=
= -2sen[(a + a)/2] × 1 × (1/2)
= -sina
Pregunta 6. lím x→π/4 [(1 – tanx)/(x – π/4)]
Solución:
Tenemos,
lím x→π/4 [(1 – tanx)/(x – π/4)]
Consideremos, y = [x – π/4]
Aquí, x→π/4, y→0
=
=
=
=
=
= -2 × 1 × [1/(1 – 0)]
= -2
Pregunta 7. lím x→π/2 [(1 – senx)/(π/4 – x) 2 ]
Solución:
Tenemos,
lím x→π/2 [(1 – senx)/(π/4 – x) 2 ]
Consideremos, y = [π/2 – x]
Aquí, x→π/2, y→0
=
= lim y→0 [(1 – acogedor)/y 2 ]
=
= 2 × 1 × (1/4)
= (1/2)
Pregunta 8. lím x→π/3 [(√3 – tanx)/(π – 3x)]
Solución:
Tenemos,
lím x→π/3 [(√3 – tanx)/(π – 3x)]
Consideremos, y = [π/3 – x]
Cuando, x→π/3, y→0
=
=
=
=
=
= (4/3) × 1 × [1/(1 + 0)]
= (4/3)
Pregunta 9. lim x→a [(asinx – xsina)/(ax 2 – xa 2 )]
Solución:
Tenemos,
lím x→a [(asinx – xsina)/(ax 2 – xa 2 )]
= lím x→a [(asenx – xsina)/{ax(x – a)}]
Consideremos, y = [x – a]
Cuando, x→a, y→0
= lim y→0 [{asin(y + a) – (y + a)sina)}/{a(y + a)y}]
= lim y→0 [(a.seny.cosa + asina.cosy – ysina – asina)/{a(y + a)y}]
= lim y→0 [{a.seny.cosa + a.sina.(cosy – 1) – y.sina}/{a(y + a)y}]
= lim y→0 [{a.seny.cosa + a.sina.2sen 2 (y/2) – t.sina}/{a(y + a)y}]
= lim y→0 [a.seny.cosa/a(y + a)y] – lim y→0 [2.a.sina.sen 2 (y/2)/a(y + a)y] + lim y→0 [y.sina/a(a + y)y]
= [(a.cosa)/a 2 ] – [(sina)/a 2 ] + 0
= [(a.cosa – sina)/a 2 ]
Pregunta 10. lím x→π/2 [{√2 – √(1 + senx)}/cos 2 x]
Solución:
Tenemos,
lím x→π/2 [{√2 – √(1 + senx)}/cos 2 x]
Al racionalizar el numerador, obtenemos
= lím x→π/2 [{2 – (1 + senx)}/cos 2 x{√2 + √(1 + senx)}]
= lím x→π/2 [(1 – senx)/(1 – sen 2 x){√2 + √(1 + senx)}]
= lím x→π/2 [(1 – senx)/(1 – senx)(1 + senx){√2 + √(1 + senx)}]
= lím x→π/2 [1/(1 + senx){√2 + √(1 + senx)}]
= 1/{(1 + 1)(√2 + √2)}
= 1/4√2
Pregunta 11. lím x→π/2 [{√(2 – senx) – 1}/(π/2 – x) 2 ]
Solución:
Tenemos,
lím x→π/2 [{√(2 – senx) – 1}/(π/2 – x) 2 ]
Consideremos, y = [π/2 – x]
Aquí, x→π/2, y→0
=
= lim y→0 [{√(2 – acogedor) – 1}/y 2 ]
Al racionalizar el numerador, obtenemos
= lim y→0 [{(2 – cómodo) – 1}/y 2 {√(2 – cómodo) – 1}]
= lim y→0 [{1 – acogedor}/y 2 {√(2 – acogedor) – 1}]
=
=
= 2/4(1 + 1)
= 1/4
Pregunta 12. lím x→π/4 [(√2 – cosx – senx)/(π/4 – x) 2 ]
Solución:
Tenemos,
lím x→π/4 [(√2 – cosx – senx)/(π/4 – x) 2 ]
=
=
=
=
= 2√2/4
= (1/√2)
Pregunta 13. lím x→π/8 [(cot4x – cos4x)/(π – 8x) 3 ]
Solución:
Tenemos,
lím x→π/8 [(cot4x – cos4x)/(π – 8x) 3 ]
= lím x→π/8 [(cot4x – cos4x)/8 3 (π/8 – x) 3 ]
Consideremos, (π/8 – x) = y
Cuando x→π/8, y→0
=
= lím x→0 [(tan4x-sin4x)/8 3 (π/8-x) 3 ]
= lím x→0 [(sen4x/cos4x-sen4x)/8 3 (π/8-x) 3 ]
=
=
=
=
=
= (2 × 4 × 1 × 4 × 1)/(8 3 )
= 1/16
Pregunta 14. lim x→a [(cosx – cosa)/(√x – √a)]
Solución:
Tenemos,
lím x→a [(cosx – cosa)/(√x – √a)]
=
Al racionalizar el denominador, obtenemos
=
=
= -2 × sina × 1 × (1/2) × 2√a
= -2√a.sina
Pregunta 15. lim x→π [{√(5 + cosx) – 2}/(π – x) 2 ]
Solución:
Tenemos,
límite x→π [{√(5 + cosx) – 2}/(π – x) 2 ]
Consideremos, y = [π – x]
Cuando, x→π, y→0
=
= lim y→0 [{√(5 – acogedor) – 2}/y 2 ]
Al racionalizar el numerador, obtenemos
=
= lim y→0 [{1 – acogedor}/y 2 {√(5 – acogedor)-2}]
=
=
= 2 × (1/4) × {1/(2 + 2)}
= (1/8)
Pregunta 16. lím x→a [(cos√x – cos√a)/(x – a)]
Solución:
Tenemos,
lím x→a [(cos√x – cos√a)/(x – a)]
=
=
= -2sin√a × 1 × (1/2√a) × (1/2)
= -(sen√a/2√a)
Pregunta 17. lim x→a [(sin√x – sin√a)/(x – a)]
Solución:
Tenemos,
lím x→a [(sin√x – sin√a)/(x – a)]
=
=
= 2cos√a × 1 × (1/2√a) × (1/2)
= (cos√a/2√a)
Pregunta 18. lím x→1 [(1 – x 2 )/sin2πx]
Solución:
Tenemos,
límite x→1 [(1 – x 2 )/sen2πx]
Cuando, x→1, h→0
= lím h→0 [{1-(1-h) 2 }/sen2π(1-h)]
= lím h→0 [(2h-h 2 )/-sen2πh]
= lím h→0 [{h(2-h)}/sen2πh]
=
=
= -2/2π
= -1/π
Pregunta 19. lím x→π/4 [{f(x) – f(π/4)}/{x – π/4}]
Solución:
Tenemos,
límite x→π/4 [{f(x) – f(π/4)}/{x – π/4}]
Cuando, x→π/4, h→0
= lím h→0 [{f(π/4 + h) – f(π/4)}/{π/4 + h – π/4}]
Se da que f(x) = sen2x
= lím h→0 [{sin(π/2 + 2h) – sin(π/2)}/h]
= lím h→0 [(cos2h – 1)/h]
= lím h→0 [{-2sen 2 h}/h]
= -2Lim h→0 [(senh/h) 2 ] × h
= -2 × 1 × 0
= 0
Publicación traducida automáticamente
Artículo escrito por vivekray59 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA