1 Introduction

Oftentimes in calculus, we must compute limits that involve trig functions. For example, $$\lim_{x\to{a}}\sin{x}$$ or the limit of any other expression that contains a trig function. This specific limit asks “what value does $\sin{x}$ approach as $x$ approaches $a$?”

Keep in mind that some trig limits do not exist. Because of the oscillation of basic trig functions it is impossible to find the limit as these functions go to infinity or negative infinity.

2 Key Formulas

There are some key formulas that you can apply when computing trig limits as they approach $0$ which is a very frequently asked question. They are as follows: $$\lim_{x\to{0}}\dfrac{\sin{x}}{x}=1.$$ And, $$\lim_{x\to{0}}\dfrac{\cos{x}-1}{x}=0.$$
Using limit rules, it is often possible to rewrite limits in terms of these common expressions and make substitutions.

3 Using Direct Substitution

Trig limits can sometimes be evaluated using direct substitution. As always this is the first approach that should be tried when computing limits. Given a limit $$\lim_{x\to{a}}f(x)$$ substitute $a$ and compute $f(a)$.

4 Example Problem

Let’s look at an example using direct substitution and basic trig rules.
Suppose we wish to evaluate $$\lim_{x\to{0}}\dfrac{5\cos{x}\sin{x}}{6x}.$$ Initially this problem looks complicated, however, we can break down this limit into multiple parts and compute the limit:

\[ \begin{align*} \lim_{x\to{0}}\dfrac{5\cos{x}\sin{x}}{6x} & =\lim_{x\to{0}}\dfrac{5}{6}\cdot\lim_{x\to{0}}\cos{x}\cdot\lim_{x\to{0}}\dfrac{\sin{x}}{x} & = \dfrac{5}{6}\cdot1\cdot1 & = \dfrac{5}{6}. \end{align*} \]

5 Conclusion

As you can see using a mix of direct substitution and key formulas we can compute what appear to be difficult problems relatively easily. You should now be comfortable with breaking down trig limits into pieces and using key formulas and direct substitution to evaluate them.