Find the limit. lim _(xarrow 0)(tan(12x))/(sin(11x))= square

To find the limit of the given expression, we can use the fact that $\tan(12x) = \frac{\sin(12x)}{\cos(12x)}$. Then, we can apply the limit to the numerator and denominator separately.<br /><br />$\lim _{x\rightarrow 0}\frac {\tan(12x)}{\sin(11x)} = \lim _{x\rightarrow 0}\frac {\frac{\sin(12x)}{\cos(12x)}}{\sin(11x)}$<br /><br />Now, we can apply the limit to the numerator and denominator separately:<br /><br />$\lim _{x\rightarrow 0}\frac {\frac{\sin(12x)}{\cos(12x)}}{\sin(11x)} = \lim _{x\rightarrow 0}\frac {\sin(12x)}{\cos(12x)\sin(11x)}$<br /><br />As $x$ approaches 0, $\sin(12x)$ and $\sin(11x)$ both approach 0, while $\cos(12x)$ approaches 1. Therefore, the limit of the expression is:<br /><br />$\lim _{x\rightarrow 0}\frac {\sin(12x)}{\cos(12x)\sin(11x)} = \frac {0}{1 \cdot 0} = \frac {0}{0}$<br /><br />This is an indeterminate form, so we need to use L'Hopital's rule. L'Hopital's rule states that if the limit of the numerator and denominator both approach 0 or both approach infinity, we can take the derivative of the numerator and denominator separately and then find the limit.<br /><br />Taking the derivative of the numerator and denominator with respect to $x$, we get:<br /><br />$\frac{d}{dx}(\sin(12x)) = 12\cos(12x)$<br /><br />$\frac{d}{dx}(\cos(12x)\sin(11x)) = -12\sin(12x)\cos(12x) + 11\cos(12x)\sin(11x)$<br /><br />Now, we can apply L'Hopital's rule:<br /><br />$\lim _{x\rightarrow 0}\frac {\sin(12x)}{\cos(12x)\sin(11x)} = \lim _{x\rightarrow 0}\frac {12\cos(12x)}{-12\sin(12x)\cos(12x) + 11\cos(12x)\sin(11x)}$<br /><br />As $x$ approaches 0, $\cos(12x)$ and $\sin(11x)$ both approach 1, while $\sin(12x)$ approaches 0. Therefore, the limit of the expression is:<br /><br />$\lim _{x\rightarrow 0}\frac {12\cos(12x)}{-12\sin(12x)\cos(12x) + 11\cos(12x)\sin(11x)} = \frac {12 \cdot 1}{-12 \cdot 0 + 11 \cdot 1} = \frac {12}{11}$<br /><br />Therefore, the limit of the given expression is $\frac {12}{11}$.