Question

(xtan(x)/sin(3x))limx=0​

Answer 1

Question :

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)}}$

Solution :

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)}}$

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{(0)tan(0)}{sin(3 \times 0)}}$

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x))}{sin(3x)} = \dfrac{0}{0}}$

Now, by using the L'Hopital's rule, we get :

$\underline{\sf{\bigstar\: L'Hopital's\:rule :-}} \\ \\ \sf{:\implies \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}}$

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{\dfrac{d}{dx}\left[(x)tan(x)\right]}{\dfrac{d}{dx}\left[sin(3x)\right]}}$

Now,

• Derivative of (x)tan(x) :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d}{dx}\left[(x)tan(x)\right]}$

By using the product rule of differentiation, we get :

$\underline{\sf{\bigstar\: Product\:rule\:of\: differentiation :-}} \\ \\ \sf{:\implies \dfrac{d(uv)}{dx} = u\cdot\dfrac{d(v)}{dx} + v\cdot\dfrac{du}{dx}}$

Here,

• u = x
• v = tan(x)

$:\implies \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)\cdot\dfrac{d[tan(x)]}{dx} + tan(x)\cdot\dfrac{d(x)}{dx}}$

$:\implies \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)\cdot sec^{2}(x) + tan(x)\cdot1}$

$:\implies \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)sec^{2}(x) + tan(x)}$

$\boxed{\therefore \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)sec^{2}(x) + tan(x)}}$

Hence, the derivative of (x)tan(x) is (x)sec²(x) + tan(x).

• Derivative of sin(3x) :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d}{dx}\left[sin(3x)\right]}$

By using the chain rule of differentiation, we get :

$\underline{\sf{\bigstar\: Chain\:rule\:of\: differentiation :-}} \\ \\ \sf{:\implies \dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

Here,

• u = 3x
• v = sin(3x)

$:\implies \sf{\dfrac{d[sin(3x)]}{dx} = \dfrac{d[sin(3x)]}{d(3x)}\cdot\dfrac{d(3x)}{dx}}$

$:\implies \sf{\dfrac{d[sin(3x)]}{dx} = cos(3x)\cdot3}$

$:\implies \sf{\dfrac{d[sin(3x)]}{dx} = 3cos(3x)}$

$\boxed{\therefore \sf{\dfrac{d[sin(3x)]}{dx} = 3cos(3x)}}$

Hence, the derivative of sin(3x) is 3cos(3x).

Now, by substituting the derivative of (x)tan(x) and sin(3x) in the equation, we get :

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{\dfrac{d}{dx}\left[(x)tan(x)\right]}{\dfrac{d}{dx}\left[sin(3x)\right]}}$

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{(x)sec^{2}(x) + tan(x)}{3cos(3x)}}$

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \dfrac{(0)sec^{2}(0) + tan(0)}{3cos(3(0))}}$

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \dfrac{0}{1}}$

$:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = 0}$

$\boxed{\therefore \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = 0}}$

Answer 2

Answer:

$Question :</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)</p><p></p><p>Solution :</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{(0)tan(0)}{sin(3 \times 0)}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)=x→0limsin(3×0)(0)tan(0)</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x))}{sin(3x)} = \dfrac{0}{0}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x))=00</p><p></p><p>Now, by using the L'Hopital's rule, we get :</p><p></p><p>\begin{gathered}\underline{\sf{\bigstar\: L'Hopital's\:rule :-}} \\ \\ \sf{:\implies \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}} \\ \\ \end{gathered}★L′Hopital′srule:−:⟹x→alimg(x)f(x)=x→alimg′(x)f′(x)</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{\dfrac{d}{dx}\left[(x)tan(x)\right]}{\dfrac{d}{dx}\left[sin(3x)\right]}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)=x→0limdxd[sin(3x)]dxd[(x)tan(x)]</p><p></p><p>Now,</p><p></p><p>Derivative of (x)tan(x) :</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d}{dx}\left[(x)tan(x)\right]} \\ \\ \end{gathered}:⟹dxdy=dxd[(x)tan(x)]</p><p></p><p>By using the product rule of differentiation, we get :</p><p></p><p>\begin{gathered}\underline{\sf{\bigstar\: Product\:rule\:of\: differentiation :-}} \\ \\ \sf{:\implies \dfrac{d(uv)}{dx} = u\cdot\dfrac{d(v)}{dx} + v\cdot\dfrac{du}{dx}} \\ \\ \end{gathered}★Productruleofdifferentiation:−:⟹dxd(uv)=u⋅dxd(v)+v⋅dxdu</p><p></p><p>Here,</p><p></p><p>u = x</p><p></p><p>v = tan(x)</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)\cdot\dfrac{d[tan(x)]}{dx} + tan(x)\cdot\dfrac{d(x)}{dx}} \\ \\ \end{gathered}:⟹dxd[(x)tan(x)]=(x)⋅dxd[tan(x)]+tan(x)⋅dxd(x)</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)\cdot sec^{2}(x) + tan(x)\cdot1} \\ \\ \end{gathered}:⟹dxd[(x)tan(x)]=(x)⋅sec2(x)+tan(x)⋅1</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)sec^{2}(x) + tan(x)} \\ \\ \end{gathered}:⟹dxd[(x)tan(x)]=(x)sec2(x)+tan(x)</p><p></p><p>\begin{gathered}\boxed{\therefore \sf{\dfrac{d[(x)tan(x)]}{dx} = (x)sec^{2}(x) + tan(x)}} \\ \\ \end{gathered}∴dxd[(x)tan(x)]=(x)sec2(x)+tan(x)</p><p></p><p>Hence, the derivative of (x)tan(x) is (x)sec²(x) + tan(x).</p><p></p><p>Derivative of sin(3x) :</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d}{dx}\left[sin(3x)\right]} \\ \\ \end{gathered}:⟹dxdy=dxd[sin(3x)]</p><p></p><p>By using the chain rule of differentiation, we get :</p><p></p><p>\begin{gathered}\underline{\sf{\bigstar\: Chain\:rule\:of\: differentiation :-}} \\ \\ \sf{:\implies \dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}} \\ \\ \end{gathered}★Chainruleofdifferentiation:−:⟹dxdy=dudy⋅dxdu</p><p></p><p>Here,</p><p></p><p>u = 3x</p><p></p><p>v = sin(3x)</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{d[sin(3x)]}{dx} = \dfrac{d[sin(3x)]}{d(3x)}\cdot\dfrac{d(3x)}{dx}} \\ \\ \end{gathered}:⟹dxd[sin(3x)]=d(3x)d[sin(3x)]⋅dxd(3x)</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{d[sin(3x)]}{dx} = cos(3x)\cdot3} \\ \\ \end{gathered}:⟹dxd[sin(3x)]=cos(3x)⋅3</p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{d[sin(3x)]}{dx} = 3cos(3x)} \\ \\ \end{gathered}:⟹dxd[sin(3x)]=3cos(3x)</p><p></p><p>\begin{gathered}\boxed{\therefore \sf{\dfrac{d[sin(3x)]}{dx} = 3cos(3x)}} \\ \\ \end{gathered}∴dxd[sin(3x)]=3cos(3x)</p><p></p><p>Hence, the derivative of sin(3x) is 3cos(3x).</p><p></p><p>Now, by substituting the derivative of (x)tan(x) and sin(3x) in the equation, we get :</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{\dfrac{d}{dx}\left[(x)tan(x)\right]}{\dfrac{d}{dx}\left[sin(3x)\right]}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)=x→0limdxd[sin(3x)]dxd[(x)tan(x)]</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \lim_{x \to 0} \dfrac{(x)sec^{2}(x) + tan(x)}{3cos(3x)}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)=x→0lim3cos(3x)(x)sec2(x)+tan(x)</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \dfrac{(0)sec^{2}(0) + tan(0)}{3cos(3(0))}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)=3cos(3(0))(0)sec2(0)+tan(0)</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = \dfrac{0}{1}} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)=10</p><p></p><p>\begin{gathered}:\implies \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = 0} \\ \\ \end{gathered}:⟹x→0limsin(3x)(x)tan(x)=0</p><p></p><p>\begin{gathered}\boxed{\therefore \sf{\lim_{x \to 0} \dfrac{(x)tan(x)}{sin(3x)} = 0}} \\ \\ \end{gathered}∴x→0limsin(3x)(x)tan(x)=0</p><p></p><p>$