Question

limitxtends to 0 tan 8x/sin2x​

Answer 1

Answer:

don't know............................

Answer 2

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \large \pink{ \bold{ \underline{ step \: by \: step \: explaination}}}$

$\quad\implies\tt{\lim_{x \to 0} \dfrac{tan8x}{sin2x}}$

$\quad\implies\bf{\bullet \: Properties }$

$\clubsuit\bf{\lim_{x \to 0} \dfrac{tanx}{x} =1}$

$\clubsuit\bf{\lim_{x \to 0} \dfrac{sinx}{x}=1}$

$\quad\implies\bf{\bullet \: Apply \: the \: properties }$

$\quad\implies\displaystyle\sf{\lim_{x \to 0} \dfrac{tan8x}{sin2x}}$

$\quad\implies\displaystyle\sf{\lim_{x \to 0} \dfrac{\dfrac{tan8x \times 8x}{8x}}{\dfrac{\sin2x \times 2x}{2x}}</p><p>}$

$\quad\implies\displaystyle\sf{\lim_{x \to 0} \dfrac{8x}{2x}}$

$\quad\implies\displaystyle\sf{\lim_{x \to 0}\cancel\dfrac{8x}{2x}}$

$\quad\implies\displaystyle\sf{\lim_{x \to 0} \: 4}$

$\boxed{\underline{\bf{\lim_{x \to 0} \dfrac{tan8x}{sin2x}=4}}}$