Question

Evaluate the limit. Be sure to show your reasoning. limit as x approaches 0 (sin7x / tan3x) ?

I know that the bottom turns to sin3x/cos3x and then i multiply the recipricol in which I get the limit as x approaches 0 (sin x7)(cos3x / sin 3x) but I don't know where to go from here :(

Answer 1

`7/3`

Explanation:

`sin(7x)/tan(3x)`
`=((7x)+(7x)^3/3!...)/((3x)+(1/3)(3x)^3)...`(Maclaurin series)
`=(7+O(x^2))/(3+O(x^2))` (canceling `x`'s throughout)
`to 7/3` as `x to 0`

Answer 2

`7/3`

Explanation:

use L'Hôpital's Rule (also known as hospital's rule in English)
https://www.google.co.in/search?q=lhopital+rule&ie=&oe=

what this rule basically means is, if we have a limit for the division of two functions which looks like `0/0` or `oo/oo` which are indeterminate forms, we can divide the derivatives of the two functions to get the answer of the limit

therefore,
`lim_(x->c) f(x)/g(x) = lim_(x->c) (f'(x))/(g'(x))` if f(x) and g(x) would both be 0 or infinity or negative infinity if evaluated at c

therefore, taking the derivative of both the functions, we get
`lim_(x->0) sin(7x)/tan(3x) =lim_(x->0) (d/dx sin(7x))/(d/dx tan(3x))`
=

which is equal to

`lim_(x->0) (7cos(7x)) /(3 sec^2 (3x))`

now since `x` goes to 0 ,`7x` and `3x` would also approach 0

= `(7cos(0)) /(3 sec^2 (0))`

and `cos(0)` and `sec^2(0)` are both 1

therefore, the limit is
`7/3`
-

Answer 3

Without using derivatives, see below.

Explanation:

As you have said:

`(sin7x)/(tan3x) = (sin7x)/((sin3x)/(cos3x))`

`= (sin7x)/1 1/(sin3x) (cos3x)/1`

Now we'll write this so we can use
`lim_(trarr0)sint/t=1` and `lim_(trarr0)1/sint=1`

`= (sin7x)/(7x) * (7x)/(3x) (3x)/(sin3x) (cos3x)/1`

`=7/3 (sin7x)/(7x) (3x)/(sin3x) (cos3x)/1`

Now take limit as `xrarr0` (so both `3x` and `7x` also go to `0`)

`lim_(xrarr00)(sin7x)/(tan3x) = lim_(xrarr0)(7/3 (sin7x)/(7x) (3x)/(sin3x) (cos3x)/1`

`= 7/3 (1) (1) 1/1 = 7/3`