LIMITS EXAMPLES AND SOLUTIONS

Question 1 :

Evaluate the following limit 

lim _{x-> 0}  (1 – cos²x)/(x sin 2x)

Solution :

  =  lim _{x-> 0}  (sin²x)/(2x sin x cos x)

  =  lim _{x-> 0}  (1/2)(sinx)/(x cos x)

  =  (1/2) lim _{x-> 0}  (tanx/x)

  =  (1/2) (1)

  =  1/2

Question 2 :

Evaluate the following limit 

lim _{x-> ∞}  x (3^1/x + 1 – cos (1/x) – e^1/x)

Solution :

=  lim _{x-> ∞}  x (3^1/x + 1 – cos (1/x) – e^1/x)

=  lim _{x-> ∞}  (3^1/x + 1 – cos (1/x) – e^1/x)/(1/x)

let y = 1/x

If x -> ∞, then y -> 0

  =  lim  _{y-> 0}   (3^y + 1 – cos y – e^y)/y

  =  lim  _{y-> 0} (3^y + 1 – cos y – e^y + 1 - 1)/y

  =  lim  _{y-> 0}  [(3^y - 1) + (1 – cos y) - (e^y - 1)]/y

=  [lim _{y-> 0} (3^y-1)/y]+[lim  _{y-> 0} (1–cos y)/y]-[lim  _{y-> 0} (e^y-1)]/y]

=  log 3 + 0 - 1

=  log 3 - 1

Hence the value of lim _{x-> ∞}  x (3^1/x + 1 – cos (1/x) – e^1/x) is log 3 - 1.

Question 3 :

Evaluate the following limit 

lim _{x-> ∞}  {x [log (x + a) – log x] }

Solution :

  =  lim _{x-> ∞}  {x [log (x + a) – log x] }

  =  lim _{x-> ∞}  {x [log (x + a)/x] }

  =  lim _{x-> ∞}  (log (1 + a/x))/(1/x)

  =  lim _{x-> ∞}  (a/a)(log (1 + a/x))/(1/x)

  =  lim _{x-> ∞}  a(log (1 + a/x))/(a/x)

Let y = a/x

If x -> ∞, then y -> 0

  =  lim _{y -> 0}  a(log (1 + y )/y)

  =  a

Question 4 :

Evaluate the following limit 

lim _x-> π  sin 3x/sin 2x

Solution :

  =  lim _x-> π  sin 3x/sin 2x

By applying the limit in the given question, we get 0/0.

  =  lim _x-> π  sin (2x + x)/sin 2x

  =  lim _x-> π  (sin 2x cos x + cos 2x sin x)/sin 2x

  =  lim _x-> π [(sin 2x cos x)/sin 2x + (cos 2x sin x)/sin 2x]

Applying the formula for sin 2x  =  2 sin x cos x 

  =  lim _x-> π cos x + lim _x-> π (cos 2x sin x)/(2 sin x cos x)

=  cos π + cos 2π/2cos π

=  -1 - (1/2)

=  -3/2

Hence the value of lim _x-> π  sin 3x/sin 2x is -3/2.

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