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

2 Answers
Jul 14, 2016

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

Jul 14, 2016

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.