Question

Question #56924

Answer 1

`" The Limit="3/8`.

Explanation:

Let us use the following Standard Forms of Limit :

`(1) lim_(xrarr0)sinx/x=1, &, (2) lim_(xrarr0)tanx/x=1`

Hence, Reqd. Lim. `=lim_(xrarr0){(tan3x)/(3x)(3x)}/{(sin8x)/((8x))(8x))`

`=lim_(xrarr0){(tan3x)/(3x)}/{(sin8x)/(8x)}(3/8)`

`=(1/1)(3/8)`

`:." The Limit="3/8`.

Answer 2

When `lim_(xto0) tan(3x)/sin(8x)` is evaluated at 0, we obtain `0/0`; this means that L'Hôpital's rule should be used.

Explanation:

To use L'Hôpital's rule:

1. Compute the derivative of the numerator: `(d(tan(3x)))/dx = 3sec^2(3x)`

2. Compute the derivative of the denominator: `(d(sin(8x)))/dx = 8cos(8x)`

3. Write the limit with the new numerator and the new denominator: `lim_(xto0)(3sec^2(3x))/(8cos(8x))` This limit can be found by evaluation at 0:

`lim_(xto0)(3sec^2(3x))/(8cos(8x)) = 3/8`

L'Hôpital's rule states that; as this limit goes, so goes the original limit:

`lim_(xto0) tan(3x)/sin(8x) = 3/8`