Question

Lim x tends to 0 (xtanx)/(sin3x)

anser is 0

Answer 1

S O L U T I O N:

Given Limit:

$\displaystyle\rm=\lim_{x\to0}\bigg[\dfrac{x\tan(x)}{\sin(3x)}\bigg]$

If we put x = 0, we get:

$\displaystyle\rm=\dfrac{0\tan(0)}{\sin(0)}$

$\rm=\dfrac{0}{0}$

Which is an indeterminate form.

So, let us use L'Hopital's rule here.

Given Limit:

$\displaystyle\rm=\lim_{x\to0}\bigg[\dfrac{x\tan(x)}{\sin(3x)}\bigg]$

Using L'Hopital's rule, we get:

$\displaystyle\rm=\lim_{x\to0}\bigg[\dfrac{\dfrac{\cos(x)\sin(x)+x}{cos^{2}(x)}}{3\cos(3x)}\bigg]$

$\displaystyle\rm=\lim_{x\to0}\bigg[\dfrac{\cos(x)\sin(x)+x}{3\cos^{2}(x)\cos(3x)}\bigg]$

Now put x = 0, we get:

$\displaystyle\rm=\dfrac{\cos(0)\sin(0)+0}{3\cos^{2}(0)\cos(0)}$

$\displaystyle\rm=\dfrac{1\times0+0}{3\times1\times1}$

$\displaystyle\rm=\dfrac{0}{3}$

$\rm=0$

Therefore:

$\displaystyle\rm\longrightarrow\lim_{x\to0}\bigg[\dfrac{x\tan(x)}{\sin(3x)}\bigg]=0$

⊕ Which is our required answer.

L E A R N  M O R E:

$\displaystyle \rm1. \: \: \lim_{x \to0} \sin(x) = 0$

$\displaystyle \rm2. \: \: \lim_{x \to0} \cos(x) = 1$

$\displaystyle \rm3. \: \: \lim_{x \to0} \dfrac{ \sin(x) }{x} = 1$

$\displaystyle \rm4. \: \: \lim_{x \to0} \dfrac{ \tan(x) }{x} = 1$

$\displaystyle \rm5. \: \: \lim_{x \to0} \dfrac{1 - \cos(x) }{x} = 1$

$\displaystyle \rm6. \: \: \lim_{x \to0} \dfrac{ \sin^{ - 1} (x) }{x} = 1$

$\displaystyle \rm7. \: \: \lim_{x \to0} \dfrac{ \tan^{ - 1} (x) }{x} = 1$

$\displaystyle \rm8. \: \: \lim_{x \to0} \dfrac{ {e}^{x} - 1 }{x} = 1$

$\displaystyle \rm9. \: \: \lim_{x \to0} \dfrac{ {a}^{x} - 1 }{x} = \ln(a)$

$\displaystyle \rm10. \: \: \lim_{x \to0} \dfrac{ \log(1 + x) }{x} = 1$

Answer 2

Answer:

0

Step-by-step explanation:

Given limit,

$\longrightarrow \lim \limits_{x \to 0} \dfrac{x \tan x}{ \sin 3x}$

The given limit will attain (0/0) form by direct substitution. Therefore the substitution method failed to solve our limit. We have to choose another method.

Consider again,

$\longrightarrow \lim \limits_{x \to 0} \dfrac{x \tan x}{ \sin 3x}$

Multiply both numerator and denominator with 3x.

$\longrightarrow \lim \limits_{x \to 0} \dfrac{x \tan x \times 3x}{ \sin 3x \times 3x}$

$\longrightarrow \lim \limits_{x \to 0} \dfrac{3x}{ \sin 3x} \times \dfrac{ \tan x}{x} \times \dfrac{x}{3}$

${ \longrightarrow \lim \limits_{x \to 0} \left[ \dfrac{3x}{ \sin 3x} \right]\times \lim \limits_{x \to 0} \left[\dfrac{ \tan x}{x}\right] \times \lim \limits_{x \to 0}\left[ \dfrac{x}{3}\right]}$

Now using the below formulas:

• $\boxed{\lim\limits_{x\to 0} \dfrac{\sin x}{x} = 1}$
• $\boxed{\lim\limits_{x\to 0} \dfrac{\tan x}{x} = 1}$

${ \longrightarrow \left[ 1\right]\times \left[1\right] \times \lim \limits_{x \to 0}\left[ \dfrac{x}{3}\right]}$

${ \longrightarrow \left[ \dfrac{0}{3}\right]}$

${ \longrightarrow 0}$

Hence the required answer is:

$\underline {\boxed{\lim \limits_{x \to 0} \dfrac{x \tan x}{ \sin 3x} = 0}}$