Question

Evaluate:
lim x-->1 [2sin(x-1)/(x-1)]
where [.] represents greatest integer function.
please someone solve it
Answer is 1 how?​

Answer 1

Given :  $\lim_{x \to 1} \left[ \dfrac{2\sin{(x-1)}}{x-1} \right]$

[.] represents greatest integer function.

To Find :  Value

Solution:

$\lim_{x \to 1} \left[ \dfrac{2\sin{(x-1)}}{x-1} \right]$

on substituting x = 1

sin(x - 1) = sin(1 - 1) = 0

x - 1 = 1 - 1= 0

Hence its 0/0 form

so applying L' hospital rule

differentiating numerator and denominator separately

$\lim_{x \to 1} \left[ \dfrac{2\cos{(x-1)}}{1} \right]$

now cos (0) = 1  but cos (0+)  and cos(0-)  are +ve and  slightly less than 1

cos ( -α) = cos(α)

Hence cos(x- 1)  ≈ 1   < 1   as x → 1

at x = 1  Value will 2

at  x → 1 value will  1    as  [ slightly less than 2 ]  =  1

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Answer 2

SOLUTION

TO PROVE

$\displaystyle \sf{\lim_{x \to 1} \: \bigg[ \: \frac{2 \sin (x - 1)}{(x-1)} \: \bigg]}$

Where [ ] represents greatest integer function

EVALUATION

Here the given limit is

$\displaystyle \sf{\lim_{x \to 1} \: \bigg[ \: \frac{2 \sin (x - 1)}{(x-1)} \: \bigg]}$

First we explain about greatest integer function

[ x ] represents the greatest integer but not greater than x

For example

[ 2 ] = 2 , [ 1.99 ] = 1 , [ - 2 ] = - 2 , [ - 1.99 ] = - 2

Now we concentrate on the given limit

$\displaystyle \sf{\lim_{x \to 1} \: \bigg[ \: \frac{2 \sin (x - 1)}{(x-1)} \: \bigg]}$

Let h = x - 1

Then h → 0 as x → 1

Then above limit becomes

$\displaystyle \sf{\lim_{x \to 1} \: \bigg[ \: \frac{2 \sin (x - 1)}{(x-1)} \: \bigg]}$

$\displaystyle \sf{ = \lim_{h \to 0} \: \bigg[ \: \frac{2 \sin h}{h} \: \bigg]}$

$\displaystyle \sf{Now \: for \: \: 0 < h < \frac{\pi}{2} \: we \: have}$

$\displaystyle \sf{0 < \sin h < h }$

$\displaystyle \sf{ \implies \: 0 < \frac{\sin h}{h} < 1 }$

$\displaystyle \sf{ \implies \: 0 < \frac{2\sin h}{h} < 2 }$

Thus we get

Right hand limit

$\displaystyle \sf{ = \lim_{x \to 1 + } \: \bigg[ \: \frac{2 \sin (x - 1)}{(x-1)} \: \bigg]}$

$\displaystyle \sf{ = \lim_{h \to 0 + } \: \bigg[ \: \frac{2 \sin h}{h} \: \bigg]}$

$\displaystyle \sf{ = 1}$

Similarly it can be shown that

Left hand limit

$\displaystyle \sf{ = \lim_{x \to 1 - } \: \bigg[ \: \frac{2 \sin (x - 1)}{(x-1)} \: \bigg]}$

$= 1$

∴ Right hand limit = Left hand limit

$\displaystyle \sf{ \therefore \: \: \lim_{x \to 1 } \: \bigg[ \: \frac{2 \sin (x - 1)}{(x-1)} \: \bigg] = 1}$

Hence proved

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