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When working with limits that involve trigonometric functions, the key is to understand the values of sine, cosine, tangent, and their counterparts at specific points like \(\frac{\pi}{6}\). These are common angles where trigonometric values are often memorized or referenced from trigonometric tables. To evaluate a limit like \(\lim_ {x \rightarrow \frac{\pi}{6}} (\cos x + \tan x)\), we utilize these known values to simplify the calculation.

For instance, knowing that \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\), allows us to substitute directly into the equation to help find the limit. This substitution step is critical as it transforms a potentially complex limit problem into a simple arithmetic exercise.

Algebraic methods are often employed in calculus to simplify expressions and make limits more manageable. This involves manipulating the expressions using algebraic rules to find a result that otherwise might be complex.

• First, substitution is frequently used when evaluating limits. This includes replacing the variable with the value it approaches.
• Simplification is another key element, where expressions are rewritten in a more straightforward form.

In the exercise provided, after substituting \(\frac{\pi}{6}\) into the cosine and tangent functions, we faced \(\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}}\). By converting \(\frac{1}{\sqrt{3}}\) into \(\frac{\sqrt{3}}{3}\) and then finding a common denominator, we simplified the expression to \(\frac{5\sqrt{3}}{6}\). This is a prime example of using algebra to make calculus problems easier to solve.

To evaluate trigonometric functions at specific points, you need to be familiar with trigonometric values at key angles like \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), and so on.

Here’s a quick guide for \(\frac{\pi}{6}\):

• \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\)
• \(\sin \frac{\pi}{6} = \frac{1}{2}\)
• \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\)
• \(\tan \frac{\pi}{6}\) can also be written as \(\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\).

These values are critical when evaluating limits involving trigonometric functions.

For example, in the provided exercise, substituting \(\frac{\pi}{6}\) into the \(\tan x\) and \(\frac{\tan x}{y}\tan x\) functions gives us \(\frac{\tan x}{\tan x}{\frac{\tan x}}\frac{\frac{{\{\tan x}}\tan \frac{1}{\sqrt{3}}\), respectively. This simplifies the process and leads us directly to the solution.