Good work! You remembered exactly the result they're expecting you to use here! :wink:
You've got this:
. . . . .\(\displaystyle \large{\begin{array}{c}limit\\x\rightarrow 0\end{array}\,\, \frac{\sin{(x)}}{x}\, =\, 1}\)
Now they've asked you to consider this:
. . . . .\(\displaystyle \large{\begin{array}{c}limit\\x\rightarrow 0\end{array}\,\, \frac{\sin{(x)}}{3x}}\)
Rearrange the expression as follows:
. . . . .\(\displaystyle \large{\frac{\sin{(x)}}{3x}\, =\,\left(\frac{1}{3}\right)\,\left(\frac{\sin{(x)}}{x}\right)}\)
Factor the "1/3" to before the "limit", and use the previous result to find this limit's value.
Eliz.
Note: You will very likely shortly encounter things like this:
. . . . .\(\displaystyle \large{\begin{array}{c}limit\\\theta\rightarrow 0\end{array}\,\, \frac{\sin{(4\theta)}}{5\theta}}\)
This looks bad, but it works very much the same way:
. . . . .\(\displaystyle \large{\frac{\sin{(4\theta)}}{5\theta}\, =\,\left(\frac{4}{4}\right)\,\left(\frac{\sin{(4\theta)}}{5\theta}\right)\, =\,\left(\frac{4}{5}\right)\,\left(\frac{\sin{(4\theta)}}{4\theta}\right)}\)
Then you'd take the 4/5 out front, and evaluate according to the usual rule.
