Question

What is the limit as `t` approaches 0 of `(tan6t)/(sin2t)`?

Answer 1

`lim_(t->0) tan(6t)/sin(2t) = 3`. We determine this by utilising L'hospital's Rule.

To paraphrase, L'Hospital's rule states that when given a limit of the form `lim_(t→a)f(t)/g(t)`, where `f(a)` and `g(a)` are values that cause the limit to be indeterminate (most often, if both are 0, or some form of ∞), then as long as both functions are continuous and differentiable at and in the vicinity of `a,` one may state that

`lim_(t→a)f(t)/g(t)=lim_(t→a)(f'(t))/(g'(t))`

Or in words, the limit of the quotient of two functions is equal to the limit of the quotient of their derivatives.

In the example provided, we have `f(t) = tan(6t)` and `g(t)=sin(2t)`. These functions are continuous and differentiable near `t=0, tan(0) =0 and sin(0)=0`. Thus, our initial `f(a)/g(a)=0/0=?.`

Therefore, we should make use of L'Hospital's Rule. `d/dt tan(6t) = 6 sec^2(6t), d/dt sin (2t)=2 cos(2t)`. Thus...

`lim_(t->0) tan(6t)/sin(2t) = lim_(t->0) (6 sec^2(6t))/(2 cos(2t)) = (6 sec^2(0))/(2 cos(0)) = 6/(2*cos^2(0)*cos(0)) = 6/(2*1*1) = 6/2 = 3`

Answer 2

The Reqd. Lim.`=3`.

Explanation:

We will find this Limit using the following Standard Results :

`lim_(thetararr0)sintheta/theta=1, lim_(thetararr0)tantheta/theta=1`

Observe that, `tan(6t)/sin(2t)=frac(tan(6t)/(6t))(sin(2t)/(2t))` `frac(6t)(2t)=3frac(tan(6t)/(6t))(sin(2t)/(2t))`

Here, `trarr0rArr(6t)rarr0rArr lim_(trarr0)tan(6t)/(6t)=1`

Similarly, `lim_(trarr0)sin(2t)/(2t)=1`

Therefore, the Reqd. Lim.`=3{1/1}=3`.