Question

How do you find the limit of `sin(7x)/x` as x approaches 0?

Answer 1

7

Explanation:

we will use well known limit `lim_{z to 0} (sin z)/ z = 1`

so `lim_{x to 0} sin(7x)/x`

let `z = 7x` and it becomes

`lim_{z to 0} sinz/(z/7)`

`= lim_{z to 0} 7 sinz/z`

`= 7 lim_{z to 0} sinz/z = 7`

you cannot use LHopital on this <<--- the bit i would like someone else to confirm

Answer 2

`lim_(x->0) (sin 7x)/(x) = 7/1 * 1 = 7`

Explanation:

There are "two" ways to solve this limit. The first way to do is it algebraically, and the second way is to apply L'Hospital's rule (although this is technically incorrect - even though the final answer is the same).

First way:

In order to solve this limit, we'd have to use a well-known formula, namely the fact that

`lim_(x->0) (sin x)/(x) = 1`

Since we have `(sin 7x)/(x)`, we could multiply and divide by `7` to get a much nicer form, which yields

`(sin7x)/(x) = cancel(7)/1 * (sin7x)/(cancel(7)x)`

We can now rewrite our limit in the following way:

`lim_(x->0) (sin 7x)/(x) = 7/1 *lim_(x->0) (sin 7x)/(7x)`

If we let `theta = 7x`, we can then write

`lim_(x->0) (sin 7x)/(x) = 7/1 *lim_(theta->0) (sin theta)/(theta)`

Since we already know what this limit is, we simply substitute:

`lim_(x->0) (sin 7x)/(x) = 7/1 * 1 = 7`

Second way:

To apply L'Hospital's rule, we have to make sure it is an indeterminate form such as `0/0` when `x->0`. There are more indeterminate forms as well.

Since direct substitution yields

`lim_(x->0) (sin 7x)/(x) -> 0/0`, we can apply L'Hospital's rule, although we've just seen that `lim_(x->0) (sin x)/(x) = 1`, thus we have a logical fallacy. In fact, L'Hospital's rule does give you the right (final answer), although the (solution) is incorrect.

Works, but incorrect:

`cancel(lim_(x->0) (sin 7x)/(x) = lim_(x->0) (7cos 7x)/(1) = (7*cos(7*0))/(1) = 7 )`

A more detailed explanation and some more examples on L'Hospital's can be found here.