lim x → π 2 cot x − cos x ( π − 2 x ) 3 is equal to Options: A . &nbsp 1 B . &nbsp π 2 C . &nbsp 1 16 D . &nbsp 0

Answer: Option C : C lim x → π 2 cot x − cos x ( π − 2 x ) 3 Let x = π 2 + t If x → π 2 , t → 0 lim t → 0 sin t − tan t − 8 t 3 = lim t → 0 ( t − t 3 3 ! + t 5 5 ! + … ) − ( t + t 3 3 + 2 t 5 15 + … ) − 8 t 3 = 1 16 We can put x = π 2 − t and we'll get L.H.L also same. Since L.H.L = R.H.L the limit exists and is = 1 16

limx→π2cotx−cosx(π−2x)3 is equal to &#x

Discussion Forum : Limits Continuity And Differentiability

Question -

$lim x \to \frac{π}{2} \frac{cot x - cos x}{(π - 2 x)^{3}}$ is equal to

Options:

A .

1

B .

\frac{π}{2}

C .

\frac{1}{16}

D .

0

Answer: Option C
:
C

lim x \to \frac{π}{2} \frac{cot x - cos x}{(π - 2 x)^{3}}

Let

x = \frac{π}{2} + t

x \to \frac{π}{2}, t \to 0

lim t \to 0 \frac{sin t - tan t}{- 8 t^{3}}

= lim t \to 0 \frac{(t - \frac{t^{3}}{3!} + \frac{t^{5}}{5!} + \dots) - (t + \frac{t^{3}}{3} + \frac{2 t^{5}}{15} + \dots)}{- 8 t^{3}}

= \frac{1}{16}

We can put

x = \frac{π}{2} - t

and we'll get L.H.L also same.
Since L.H.L = R.H.L the limit exists and is

= \frac{1}{16}

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