Question

solve $\displaystyle\lim \limits_{x\to0}\dfrac{\sin4x}{\sin2x}}$​

Answer 1

To solve :-

$\displaystyle \rm \implies\lim \limits_{x\to0} \: \dfrac{\sin4x}{\sin2x}$

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Solution :-

If we try to solve the given limit by the method of substituting to directly substitute value of x = 0, we get :

$\leadsto\rm \dfrac{\sin 4x}{\sin 2x} = \dfrac{\sin 0}{\sin 0}=\dfrac{0}{0}$

This quantity is meaningless or indeterminate, so substitution method failed.

Let's try another method !

$\displaystyle \rm \longrightarrow\lim \limits_{x\to0} \: \dfrac{\sin4x}{\sin2x}$

Multiply both numerator and denominator with $4x$.

$\displaystyle \rm \longrightarrow\lim \limits_{x\to0} \: \dfrac{\sin4x}{\sin2x} \times \frac{4x}{4x}$

$\displaystyle \rm \longrightarrow\lim \limits_{x\to0} \: \left[\dfrac{\sin4x}{4x} \times \frac{4x}{sin \: 2x} \right]$

$\displaystyle \rm \longrightarrow\lim \limits_{x\to0} \: \left[\dfrac{\sin4x}{4x} \times \frac{2x}{sin \: 2x} \times 2\right]$

$\displaystyle \rm \longrightarrow2 \times \lim \limits_{x\to0} \: \left[\dfrac{\sin4x}{4x} \times \frac{2x}{sin \: 2x} \right]$

$\displaystyle \rm \longrightarrow2 \times \lim \limits_{x\to0} \: \left[\dfrac{\sin4x}{4x} \div \frac{ \sin 2x}{ 2x} \right]$

$\displaystyle \rm \longrightarrow2 \times \lim \limits_{x\to0} \: \left[\dfrac{\sin4x}{4x} \right] \div \left[\frac{ \sin 2x}{ 2x} \right]$

$\displaystyle \rm \longrightarrow2 \times \lim \limits_{x\to0} \: \left[\dfrac{\sin4x}{4x} \right] \div \lim \limits_{x\to0}\left[\frac{ \sin 2x}{ 2x} \right]$

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Now as we know that,

since x is approaching to 0 in both the limits, 4x and 2x will also approach to 0.

• $x\to 0 \implies 4x\to0$
• $x\to 0 \implies 2x \to 0$

[ By multiplying both sides by 4 and 2 respectively ]

We get . . .

$\displaystyle \rm \longrightarrow2 \times \lim \limits_{4x\to0} \: \left[\dfrac{\sin4x}{4x} \right] \div \lim \limits_{2x\to0}\left[\frac{ \sin 2x}{ 2x} \right]$

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Apply formula :-

$\boxed{\displaystyle\rm{\purple{\lim\limits_{x\to0}\:\left[\dfrac{sin x}{x}\right] = 1 }}}$

We get . . .

$\displaystyle \rm \longrightarrow2 \times\: \left[1 \right] \div\left[1 \right]$

$\displaystyle \rm \longrightarrow2$

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Hence,

$\displaystyle \rm \longrightarrow \underline{ \underline{\lim \limits_{x\to0} \:\left[ \dfrac{\sin4x}{\sin2x} \right]= 2}}$

Answer 2

ANSWER:

To Solve:

• $\displaystyle\lim_{x\to0}\dfrac{\sin4x}{\sin2x}$

Solution:

We are given that,

$\displaystyle\implies\lim_{x\to0}\dfrac{\sin4x}{\sin2x}$

Solving limit, by substituting the value of x as 0,

$\displaystyle \implies \dfrac{\sin4(0)}{\sin2(0)}$

$\displaystyle \implies \dfrac{\sin0}{\sin0}$

We know that sin 0 = 0.

So,

$\displaystyle \implies \dfrac{0}{0}$

As 0/0 is an indefinite form, we need to modify our fraction.

METHOD 1: USING TRIGONOMETRIC IDENTITY

So, we had,

$\displaystyle \implies \lim_{x\to0}\dfrac{\sin4x}{\sin2x}$

We know that,

$\hookrightarrow \implies \sin2\theta=2\sin\theta\cos\theta$

Using this formula in numerator,

$\displaystyle \implies \lim_{x\to0}\dfrac{\sin4x}{\sin2x}$

$\displaystyle \implies \lim_{x\to0}\dfrac{2\sin2x\cos2x}{\sin2x}$

Cancelling sin2x,

$\displaystyle \implies \lim_{x\to0}2\cos2x$

Solving limit, by substituting the value of x as 0,

$\displaystyle \implies 2\cos2(0)$

$\displaystyle \implies 2\cos0$

We know that cos 0 = 1

$\displaystyle \implies 2(1)$

$\displaystyle\bf \implies 2$

METHOD 2: USING SPECIAL FORMULAE

So, we had,

$\displaystyle \implies \lim_{x\to0}\dfrac{\sin4x}{\sin2x}$

Multiplying and dividing 2x and 4x both,

$\displaystyle \implies \lim_{x\to0}\dfrac{\sin4x}{\sin2x}\times\dfrac{2x}{2x}\times\dfrac{4x}{4x}$

On rearranging,

$\displaystyle \implies \lim_{x\to0}\dfrac{\sin4x}{4x}\times\dfrac{2x}{\sin2x}\times\dfrac{4x}{2x}$

We know that,

$\hookrightarrow \lim_{h\to0}\dfrac{\sin h}{h} = 1$

So,

$\displaystyle \implies \lim_{x\to0}\dfrac{\sin4x}{4x}\times\dfrac{2x}{\sin2x}\times2$

Solving limit,

$\displaystyle \implies 1\times1\times2$

$\displaystyle\bf \implies 2$

Hence,

$\displaystyle\bf\implies\lim_{x\to0}\dfrac{\sin4x}{\sin2x}=2$