Question

Someone, please solve this equation without using L'Hopital's rule.7D0%5C%20%20%5Cfrac%7Bsin%5E%7B2%7D%286x%29%20%7D%7B9x%5E%7B2%7D%20%7D" id="TexFormula1" title="\lim_{x \to \0}0\ \frac{sin^{2}(6x) }{9x^{2} }" alt="\lim_{x \to \0}0\ \frac{sin^{2}(6x) }{9x^{2} }" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$\displaystyle \large \boxed{\sf \bf \ \ \lim_{x\rightarrow0} {\dfrac{sin^2(6x)}{9x^2}}=4 \ \ }$

Step-by-step explanation:

Hello, sin(x) is equivalent of x when x tends to 0, so

$\displaystyle \lim_{x\rightarrow0} {\dfrac{sin^2(6x)}{9x^2}}\\\\=\lim_{x\rightarrow0} {\dfrac{(6x)^2}{9x^2}}\\\\=\lim_{x\rightarrow0} {\dfrac{36x^2}{9x^2}}\\\\=\lim_{x\rightarrow0} {\dfrac{36}{9}}\\\\=4$

Thank you.