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High School Math : Understanding the Definition of Limits

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Example Question #1 : Limits

$f(x)=\frac{x^{2}-x-6}{x+2}$ $\displaystyle f(x)=\frac{x^{2}-x-6}{x+2}$

$\lim_{x\rightarrow -2}f(x) =$ $\displaystyle \lim_{x\rightarrow -2}f(x) =$

Possible Answers:

$\displaystyle 0$

$\displaystyle 2$

$\displaystyle -5$

$\displaystyle -3$

$\displaystyle 6$

Correct answer:

$\displaystyle -5$

Explanation:

A limit describes what $\displaystyle y$-value a function approaches as $\displaystyle x$ approaches a certain value (in this case, $\displaystyle -2$). The easiest way to find what $\displaystyle y$-value a function approaches is to substitute the $\displaystyle x$-value into the equation.

$f(-2)=\frac{(-2)^{2}-(-2)-6}{-2+2}=\frac{4+2-6}{0}=\frac{0}{0}$ $\displaystyle f(-2)=\frac{(-2)^{2}-(-2)-6}{-2+2}=\frac{4+2-6}{0}=\frac{0}{0}$

Substituting $\displaystyle -2$ for $\displaystyle x$ gives us an undefined value (which is NOT the same thing as 0). This means the function is not defined at that point. However, just because a function is undefined at a point doesn't mean it doesn't have a limit. The limit is simply whichever value the function is getting close to.

One method of finding the limit is to try and simplify the equation as much as possible:

$f(x)=\frac{(x-3)(x+2)}{x+2}$ $\displaystyle f(x)=\frac{(x-3)(x+2)}{x+2}$

As you can see, there are common factors between the numerator and the denominator that can be canceled out. (Remember, when you cancel out a factor from a rational equation, it means that the function has a hole -- an undefined point -- where that factor equals zero.)

After canceling out the common factors, we're left with:

$\displaystyle f(x)=x-3$

Even though the domain of the original function is restricted ($\displaystyle x$ cannot equal $\displaystyle -2$), we can still substitute into this simplified equation to find the limit at x=-2. $\displaystyle x=-2.$

$\displaystyle f(-2)=-2-3=-5$

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High School Math : Understanding the Definition of Limits

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Example Question #1 : Limits

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