Using lim _(Theta arrow 0)(sinTheta )/(Theta )=1 find the limit. lim _(xarrow 0)(9sin(sin9x))/(sin9x) Select the correct answer below and, if necessary,fill in the answer box to complete your choice. A. lim _(xarrow 0)(9sin(sin9x))/(sin9x)= (Simplify your answer.) B. The limit does not exist

To find the limit $\lim _{x\rightarrow 0}\frac {9sin(sin9x)}{sin9x}$, we can use the given limit $\lim _{\Theta \rightarrow 0}\frac {sin\Theta }{\Theta }=1$.<br /><br />Let's rewrite the expression as:<br />$\lim _{x\rightarrow 0}\frac {9sin(sin9x)}{sin9x} = 9 \cdot \lim _{x\rightarrow 0}\frac {sin(sin9x)}{sin9x}$<br /><br />Now, let $\Theta = sin9x$. As $x$ approaches 0, $\Theta$ also approaches 0. So, we can substitute $\Theta$ for $sin9x$ in the limit:<br /><br />$9 \cdot \lim _{\Theta \rightarrow 0}\frac {sin\Theta }{sin9x}$<br /><br />Since $\Theta$ approaches 0 as $x$ approaches 0, we can use the given limit $\lim _{\Theta \rightarrow 0}\frac {sin\Theta }{\Theta }=1$:<br /><br />$9 \cdot \lim _{\Theta \rightarrow 0}\frac {sin\Theta }{\Theta } \cdot \lim _{\Theta \rightarrow 0}\frac {\Theta }{sin9x}$<br /><br />The second limit is equal to 1 as $\Theta$ approaches 0:<br /><br />$9 \cdot 1 \cdot \lim _{\Theta \rightarrow 0}\frac {\Theta }{sin9x}$<br /><br />Now, we need to find the limit of $\lim _{\Theta \rightarrow 0}\frac {\Theta }{sin9x}$. As $\Theta$ approaches 0, $sin9x$ also approaches 0. So, we can use the given limit $\lim _{\Theta \rightarrow 0}\frac {sin\Theta }{\Theta }=1$ again:<br /><br />$\lim _{\Theta \rightarrow 0}\frac {\Theta }{sin9x} = \lim _{\Theta \rightarrow 0}\frac {\Theta }{9\Theta } = \frac {1}{9}$<br /><br />Therefore, the limit is:<br /><br />$9 \cdot 1 \cdot \frac {1}{9} = 1$<br /><br />So, the correct answer is:<br />A. $\lim _{x\rightarrow 0}\frac {9sin(sin9x)}{sin9x} = 1$