Respuesta :

When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.

When we are attempting limits questions, there are several tests we attempt first.

1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)
[tex]\lim_{x \to 0} (\frac{sinx}{x}) = 1[/tex]
[tex]\lim_{x \to 0} (\frac{tanx}{x}) = 1[/tex]
4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.

For example:

1) [tex]\lim_{x \to 0}\frac{\sqrt{x} - 5}{x - 25}[/tex]

We can do this using the first and second method.
Method 1: Direct evaluation:

Substitute x = 0 to the function.
[tex]\frac{\sqrt{0} - 5}{0 - 25}[/tex]
[tex]= \frac{-5}{-25}[/tex]
[tex]= \frac{1}{5}[/tex]

Method 2: Rearranging the function

We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.

[tex]\lim_{x \to 0}\frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}[/tex]
[tex]= \lim_{x \to 0}\frac{1}{(\sqrt{x} + 5)}}[/tex]
[tex]= \frac{1}{5}[/tex]

Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
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