Question

39-50 find the limit.
41.
`\lim _(t \rightarrow 0) (\tan 6 t)/(\sin 2 t)`

Answer 1

Write tan in terms of sin and cos.

`\displaystyle \lim_(t\to0)(\tan(6t))/(\sin(2t)) = \lim_(t\to0)(\sin(6t))/(\sin(2t)\cos(6t))`

Recall that

`\displaystyle \lim_(x\to0)\frac{\sin(x)}x = 1`

Rewrite and expand the given limand as the product

`\displaystyle \lim_(t\to0)(\sin(6t))/(\sin(2t)\cos(6t)) = \lim_(t\to0) (\sin(6t))/(6t) * (2t)/(\sin(2t)) * (6t)/(2t\cos(6t)) \\\\ = \left(\lim_(t\to0) (\sin(6t))/(6t)\right) * \left(\lim_(t\to0)(2t)/(\sin(2t))\right) * \left(\lim_(t\to0)(3)/(\cos(6t))\right)`

Then using the known limit above, it follows that

`\displaystyle \left(\lim_(t\to0) (\sin(6t))/(6t)\right) * \left(\lim_(t\to0)(2t)/(\sin(2t))\right) * \left(\lim_(t\to0)(3)/(\cos(6t))\right) = 1 * 1 * \frac3{\cos(0)} = \boxed{3}`