# Question #21638

Jan 15, 2015

When determining the limit of a function, the first thing you should do is replace the variable with the limit itself and see if an answer exists.

In some cases, the denominator becomes 0, indicating a solution does not exist AT the limit. At this point, you proceed with the process of finding the limit (graphing, incremental approach, etc.)

However, in our specific case of:

${\lim}_{x \to 0} \frac{\sin {x}^{4}}{{\sin}^{2} \left({x}^{2}\right)}$

we actually find that:

VVVVVV incorrect from here, sorry! - Jake VVVVVV

$\sin \left({0}^{4}\right) = \sin \left(0\right) = 1$

and:

${\sin}^{2} \left({0}^{2}\right) = {\sin}^{2} \left(0\right) = 1$

Indicating that:

${\lim}_{x \to 0} \frac{\sin {x}^{4}}{{\sin}^{2} \left({x}^{2}\right)} = \frac{1}{1} = 1$

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

A first useful step in solving finite limits is always to try and substitute the value at which you are computing the limit. This can be done for infinite limits too, but infinity is thorny and we can't speak of a mere "substitution" because infinity is not a (real) number and thereby infinity's algebra has to be handled with kid gloves.

If only continuous functions are involved (continuous at the point at which you're computing the limit), for finite limits you'll get a finite value. If functions are not all continuous, then there can be some troubles that you have to deal with (indeterminate forms or nonexistence of the limit). For example in case of divisions, when the denominator's limit is $0$.

In our case ${\lim}_{x \to 0} \sin \frac{{x}^{4}}{{\sin}^{2} \left({x}^{2}\right)}$. We get:
$\sin \left({x}^{4}\right) = \sin \left({0}^{4}\right) = \sin \left(0\right) = 0$
${\sin}^{2} \left({x}^{2}\right) = {\sin}^{2} \left({0}^{2}\right) = {\sin}^{2} \left(0\right) = {0}^{2} = 0$
So this is the case of an indeterminate form $\frac{0}{0}$: division by zero is not allowed, so the function $\sin \frac{{x}^{4}}{{\sin}^{2} \left({x}^{2}\right)}$ doesn't exist for $x = 0$. This means that the simple substitution of the value at which we're computing the limit is not enough.

When dealing with $\sin$ function and limits, there's a very important notable limit that often helps:
${\lim}_{x \to 0} \frac{\sin x}{x} = 1$
This doesn't restrict its effectiveness in case of $\frac{\sin x}{x}$ only. In fact, if $f \left(x\right)$ is a real function of real variable such that ${\lim}_{x \to {x}_{0}} f \left(x\right) = 0$, where ${x}_{0} \in m a t h \boldsymbol{R}$, if $y = f \left(x\right)$ then we get that:
${\lim}_{x \to {x}_{0}} \sin f \frac{x}{f} \left(x\right) = {\lim}_{f \left(x\right) \to 0} \sin f \frac{x}{f} \left(x\right) = {\lim}_{y \to 0} \sin \frac{y}{y} = 1$

So if $f \left(x\right) = {x}^{4}$ and ${x}_{0} = 0$, we get:
${\lim}_{x \to 0} \sin \frac{{x}^{4}}{x} ^ 4 = 1$
Now the question is: can we work our original limit in some way to get this limit involved? We could multiply the numerator and the denominator by ${x}^{4}$ (which really is multiplying by $1$), getting:
${\lim}_{x \to 0} \frac{\sin \left({x}^{4}\right) \cdot {x}^{4}}{{\sin}^{2} \left({x}^{2}\right) \cdot {x}^{4}} = {\lim}_{x \to 0} \left(\frac{\sin \left({x}^{4}\right)}{{x}^{4}} \cdot {x}^{4} / \left\{{\sin}^{2} \left({x}^{2}\right)\right\}\right) = {\lim}_{x \to 0} \frac{\frac{\sin \left({x}^{4}\right)}{{x}^{4}}}{\frac{{\sin}^{2} \left({x}^{2}\right)}{x} ^ 4} = {\lim}_{x \to 0} \frac{\frac{\sin \left({x}^{4}\right)}{{x}^{4}}}{\sin \frac{{x}^{2}}{x} ^ 2} ^ 2 = \frac{1}{1} = 1$
So by some algebraic manipulations (just properties of fractions) we got ${\lim}_{x \to 0} \sin \frac{{x}^{4}}{x} ^ 4 = 1$ divided by ${\lim}_{x \to 0} \sin \frac{{x}^{2}}{x} ^ 2 = 1$ squared. So the result is $1$.

The graph of the function $\sin \frac{{x}^{4}}{{\sin}^{2} \left({x}^{2}\right)}$ looks like we're right:
graph{sin(x^4)/(sin(x^2))^2 [-1.5, 1.5, -1, 2]}