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

Jul 14, 2016

7

#### Explanation:

we will use well known limit ${\lim}_{z \to 0} \frac{\sin z}{z} = 1$

so ${\lim}_{x \to 0} \sin \frac{7 x}{x}$

let $z = 7 x$ and it becomes

${\lim}_{z \to 0} \sin \frac{z}{\frac{z}{7}}$

$= {\lim}_{z \to 0} 7 \sin \frac{z}{z}$

$= 7 {\lim}_{z \to 0} \sin \frac{z}{z} = 7$

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

Jul 14, 2016

${\lim}_{x \to 0} \frac{\sin 7 x}{x} = \frac{7}{1} \cdot 1 = 7$

#### 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

${\lim}_{x \to 0} \frac{\sin x}{x} = 1$

Since we have $\frac{\sin 7 x}{x}$, we could multiply and divide by $7$ to get a much nicer form, which yields

$\frac{\sin 7 x}{x} = \frac{\cancel{7}}{1} \cdot \frac{\sin 7 x}{\cancel{7} x}$

We can now rewrite our limit in the following way:

${\lim}_{x \to 0} \frac{\sin 7 x}{x} = \frac{7}{1} \cdot {\lim}_{x \to 0} \frac{\sin 7 x}{7 x}$

If we let $\theta = 7 x$, we can then write

${\lim}_{x \to 0} \frac{\sin 7 x}{x} = \frac{7}{1} \cdot {\lim}_{\theta \to 0} \frac{\sin \theta}{\theta}$

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

${\lim}_{x \to 0} \frac{\sin 7 x}{x} = \frac{7}{1} \cdot 1 = 7$

Second way:

To apply L'Hospital's rule, we have to make sure it is an indeterminate form such as $\frac{0}{0}$ when $x \to 0$. There are more indeterminate forms as well.

Since direct substitution yields

${\lim}_{x \to 0} \frac{\sin 7 x}{x} \to \frac{0}{0}$, we can apply L'Hospital's rule, although we've just seen that ${\lim}_{x \to 0} \frac{\sin x}{x} = 1$, 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:

$\cancel{{\lim}_{x \to 0} \frac{\sin 7 x}{x} = {\lim}_{x \to 0} \frac{7 \cos 7 x}{1} = \frac{7 \cdot \cos \left(7 \cdot 0\right)}{1} = 7}$

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