Question

Solve :-
$\lim\limits_{x\to \frac{3}{2}} \left[\dfrac{\sin(2x-3)}{(2x-3)}\right]$

I am unaware of any such method where x is not tending towards 0. Help me !!

Chapter-13 limits and derivatives

Answer 1

$\large\underline{\sf{Solution-}}$

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{3}{2} } \rm\: \dfrac{sin(2x - 3)}{2x - 3} }$

If we put directly the value of x, we get

$\rm \:  =  \:  \: \dfrac{sin\bigg(2 \times \dfrac{3}{2} - 3\bigg) }{2 \times \dfrac{3}{2} - 3}$

$\rm \:  =  \:  \: \dfrac{sin\bigg(3 - 3\bigg) }{3 - 3}$

$\rm \:  =  \:  \: \dfrac{sin0}{0}$

$\rm \:  =  \:  \: \dfrac{0}{0} \: which \: is \: meaningless$

To solve this

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{3}{2} } \rm\: \dfrac{sin(2x - 3)}{2x - 3} }$

we use method of Substitution

$\green{\rm :\longmapsto\:Put \: x \: = \: \dfrac{3}{2} - y, \: \: as \: x \to \: \dfrac{3}{2},\: \: y \to \: 0}$

So, above expression can be reduced to

$\rm \:  =  \:  \: \:\displaystyle\lim_{y \to 0} \: \dfrac{sin\bigg(2 \times \bigg(\dfrac{3}{2} - y\bigg) - 3 \bigg) }{2 \times \bigg(\dfrac{3}{2} - y\bigg) - 3 }$

$\rm \:  =  \:  \: \:\displaystyle\lim_{y \to 0} \: \dfrac{sin(3 - 2y - 3)}{3 - 2y - 3}$

$\rm \:  =  \:  \: \:\displaystyle\lim_{y \to 0} \: \dfrac{sin(- 2y)}{- 2y}$

We know,

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0}\dfrac{sinx}{x} = 1}}$

So, using this identity, we get

$\rm \:  =  \:  \: 1$

Hence,

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{3}{2} } \rm\: \dfrac{sin(2x - 3)}{2x - 3} = 1}$

Additional Information :-

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0}\dfrac{tanx}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0}\dfrac{sin {}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0}\dfrac{tan {}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga}}$

$\boxed{ \bf{ \: \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1}}$

Answer 2

Answer:

I will solve this math by La' Hospital's rule .

Rule :

La' Hospital's Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

Solution :

$\sf \: \displaystyle \lim_{x \to \frac{3}{2} } \: \frac{sin(2x - 3)}{2x - 3} \\ \\ = \displaystyle\lim_{x \to \frac{3}{2} } \frac{ \frac{d}{dx} \{sin(2x - 3) \}}{ \frac{d(2x - 3)}{dx} } \\ \\ = \displaystyle\lim_{x \to \frac{3}{2} } \frac{cos(2x - 3) \times \frac{d(2x - 3)}{dx} }{2 \times 1- 0} \\ \\ = \displaystyle\lim_{x \to \frac{3}{2} } \frac{cos(2x - 3) \times (2 \times 1- 0)}{2} \\ \\ = \displaystyle\lim_{x \to \frac{3}{2} } \frac{ \cancel{2} \: cos(2x - 3)}{ \cancel{2}} \\ \\ = \displaystyle\lim_{x \to \frac{3}{2} } \: cos(2x - 3) \\ \\ = cos(2 \times \frac{3}{2} - 3) \\ \\ = cos(3 - 3) \\ \\ = cos0 \\ \\ = 1$