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

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

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.