HOW TO EVALUATE LIMITS WITH TRIG FUNCTIONS

Question 1 :

Evaluate the following limit 

lim  _{x -> 0} sin³ (x/2)/ x³

Solution :

lim  _{x -> 0} sin³ (x/2)/ x³  =  lim  _{x -> 0} (sin (x/2))³/ x³

In order to match the given question with the formula, let us multiply and divide by 1/8

 =  lim  _{x -> 0} (sin (x/2))³(1/8)/ x³(1/8)

 =  (1/8) lim  _{x -> 0} (sin (x/2))³/ (x/2)³

 =  (1/8) lim  _{x -> 0} [sin (x/2) / (x/2)]³

 =  1/8 (1)

  =  1/8

Hence the value of lim  _{x -> 0} sin³ (x/2)/ x³ is 1/8.

Question 2 :

Evaluate the following limit 

lim  _{x -> 0} sin αx / sin ᵦx

Solution :

=  lim  _{x -> 0} sin αx / sin ᵦx

=  (lim  _{x -> 0} sin αx) (αx/αx) / (lim  _{x -> 0} sin ᵦx)(ᵦx/ᵦx)

= (αx/ᵦx) (lim  _{x -> 0} sin αx/αx) / (lim  _{x -> 0} sin ᵦx/ᵦx)

=  (α/ᵦ) (1)

=  α/ᵦ

Hence the value of lim  _{x -> 0} sin αx / sin ᵦx is α/ᵦ.

Question 2 :

Evaluate the following limit 

lim  _{x -> 0} tan 2x / sin 5x

Solution :

=  lim  _{x -> 0} tan 2x / sin 5x

=  lim  _{x -> 0} tan 2x ⋅ (2x/2x) / lim  _{x -> 0} sin 5x ⋅(5x/5x)

=  (2x/5x) [(lim  _{x -> 0} tan 2x/2x) / (lim  _{x -> 0} sin 5x/5x)

=  (2/5) (1/1)

=  2/5

Hence the value of lim  _{x -> 0} tan 2x / sin 5x is 2/5.

Question 3 :

Evaluate the following limit 

lim  _{α -> 0} (sin αⁿ)/ (sin α)^m

Solution :

=  lim  _{α -> 0} (sin αⁿ) ⋅ (αⁿ/αⁿ) / (sin α ⋅ (α/α))^m

=  lim  _{α -> 0} (α^n/α^m) lim  _{α -> 0} (sin αⁿ/αⁿ) /lim  _{α -> 0} (sin α /α)^m

=   lim  _{α -> 0} (α^n/α^m) 

=  lim  _{α -> 0} α^n-m

Question 4 :

Evaluate the following limit 

lim  _{x -> 0} (sin (a + x) - sin (a - x))/ x

Solution :

=  lim  _{x -> 0} (sin (a + x) - sin (a - x)) / x

sin C - sin D  =  2 cos ((C + D)/2) sin ((C - D)/2)

=  lim  _{x -> 0} (2 cos ((a+x+a-x)/2) sin (a+x-a+x)/2) / x

=  lim  _{x -> 0} (2 cos a sin x)/x

=  2 cos a  lim  _{x -> 0} (sin x/x)

=  2 cos a (1)

=  2 cos a

Hence the value lim  _{x -> 0} (sin (a + x) - sin (a - x))/ x is 2 cos a.

Question 5 :

Evaluate the following limit 

lim  _{x -> 1} sin (x - 1)/(x² + x - 2)

Solution :

= lim  _{x -> 1} sin (x - 1)/(x² + x - 2)

Factoring the denominator,

x² + x - 2 = (x + 2)(x - 1)

= lim  _{x -> 1} sin (x - 1)/(x + 2)(x - 1)

= lim  _{x -> 1} [sin (x - 1)/(x - 1)] lim  _{x -> 1} [1/(x + 2)]

Let  θ = x - 1

= lim  _{x -> 1} [sin θ/θ] lim  _{x -> 1} [1/(x + 2)]

= 1 lim  _{x -> 1} [1/(x + 2)]

Applying the limit, we get

= 1/3

Question 6 :

Evaluate the following limit 

lim  _{x -> -5} (x² + 3x - 10) / sin (x + 5)

Solution :

= lim  _{x -> -5} (x² + 3x - 10) / sin (x + 5)

Factoring the numerator,

= lim  _{x -> -5} (x + 5)(x - 2) / sin (x + 5)

= [lim  _{x -> -5} (x - 2)] [lim  _{x -> -5}  (x + 5) / sin (x + 5)]

Let  θ = x + 5

= [lim  _{x -> -5} (x - 2)] [lim  _{x -> -5} θ/sin θ]

Applying the limit, we get

= (-5 - 2)] (1)

= -7

Question 7 :

Evaluate the following limit 

lim  _{x -> 5} 2 tan (x - 5) / (x² - 6x + 5)

Solution :

= lim  _{x -> 5} 2 tan (x - 5) / (x² - 6x + 5)

Factoring the denominator, we get

= lim  _{x -> 5} 2 tan (x - 5) / (x - 5)(x - 1)

= lim  _{x -> 5} 2 [sin (x-5)/cos (x-5)] / (x - 5)(x - 1)

= 2 lim  _{x -> 5} [sin (x-5)/(x - 5)] lim  _{x -> 5} [1/cos (x - 5)](x - 1)

= 2 (1) lim  _{x -> 5} [1/cos (x - 5)](x - 1)]

= 2 lim  _{x -> 5} [1/cos (x - 5)](x - 1)]

Applying the limit, we get

= 2 lim  _{x -> 5} [1/cos (5 - 5)](5 - 1)]

= 2 [1/4cos 0]

= 2 [1/4 (1)]

= 2/4

= 1/2

Question 8 :

Evaluate the following limit 

lim  _{x -> -3} (x² + 4x + 3)/5 tan (x + 3)

Solution :

= lim  _{x -> -3} (x² + 4x + 3)/5 tan (x + 3)

Factoring the numerator, we get

= lim  _{x -> -3} (x + 1)(x + 3)/5 tan (x + 3)

= lim  _{x -> -3} (x + 1)(x + 3)/[5 sin(x + 3) / cos (x + 3)]

= lim  _{x -> -3} [(x + 3)/5sin(x + 3)] [(x + 1)/cos (x + 3)]

= (1/5) lim  _{x -> -3} [(x + 3)/sin(x + 3)] lim  _{x -> -3} [[(x + 1)/cos (x + 3)]

= (1/5) (1) (-3+1)/cos(-3+3)

= (1/5)[-2/cos 0]

= (1/5) (-2/1)

= -2/5

Question 9 :

Evaluate the following limit 

lim  _x ->0 sin 3x/(5x³ - 4x)

Solution :

= lim  _x ->0 sin 3x/(5x³ - 4x)

Multiplying the numerator and denominator by 3x, we get

= lim  _x ->0 sin 3x(3x/3x) / (5x³ - 4x)

= lim  _x ->0 (sin 3x/3x)  [3x/(5x³ - 4x)]

= lim  _x ->0 (sin 3x/3x)  lim  _x ->0[3x/(5x³ - 4x)]

= 1 lim  _x ->0[3x/x(5x² - 4)]

= lim  _x ->0[3/(5x² - 4)]

Applying the limit, we get

= [3/(5(0)² - 4)]

= 3/(-4)

= -3/4

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