Question

lim x tends to 0 xtanx/sin3x

and

lim x tends to infty 1/n {(m+1)(m+2) ... (m+n)}^1/n

Answer 1

Answer:

her is it

Explanation:

i only know the first one!!

Answer 2

$\\ \quad \dag \underline{ \frak{ \purple{1st \: Question : - }}}$

$\\ \quad \longrightarrow \quad \displaystyle \lim _{ x \rightarrow 0} \rm \dfrac{x \tan x}{ \sin 3x}$

L' Hospital's Rule

$\\ \quad \longrightarrow \quad \displaystyle \lim _{ x \rightarrow a} \rm \dfrac{f(x)}{g(x)} = \displaystyle \lim _{ x \rightarrow a} \rm \dfrac{f'(x)}{g'(x)}$

$\\ \quad \longrightarrow \quad \displaystyle \lim _{ x \rightarrow 0} \rm \dfrac{ \dfrac{d}{dx} \bigg( x \tan x \bigg)}{ \dfrac{d}{dx} \bigg(\sin 3x \bigg)}$

Calculating dy/dx of xtanx:-

• Apply product rule of differentiation.

$\\ \quad \mapsto \quad \sf \dfrac{d}{dx} \bigg(x \tan x \bigg) = (x )\times \dfrac{d}{dx} \bigg( \tan x \bigg) + (\tan x) \times \frac{d}{dx} \bigg(x \bigg)$

$\\ \quad \mapsto \quad \sf \dfrac{d}{dx} \bigg(x \tan x \bigg) = x { \sec}^{2} x + \tan x$

Calculating dy/dx of sin3x:-

• Apply Chain Rule Of Differentiation.

$\\ \quad \mapsto \quad \sf \dfrac{d}{dx} \bigg( \sin 3x\bigg) = 3 \cos 3x$

$\\ \quad \longrightarrow \quad \displaystyle \lim _{ x \rightarrow 0} \rm \dfrac{x { \sec}^{2}x + \tan x }{3 \cos 3x}$

$\\ \quad \longrightarrow \quad \sf \dfrac{(0) { \sec}^{2}(0) + \tan (0) }{3 \cos 3(0)} = \dfrac{0}{1} = \boxed{0} \bigstar$

Note:- Solution to the second question is provided in the attachment kindly refer to it.

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