$\displaystyle f(x) = \left\{ \begin{array}{ll} 2\cos(\pi x) \text{ for } x \leq -1 \\ \\ \displaystyle \frac{2}{\cos(\pi x)}\text{ for } x > -1 \end{array} \right.$

And we want to find:

$\displaystyle \lim_{x\to -1}f(x)$

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

$\displaystyle \lim_{x\to-1^-}f(x)$

Since we are approaching from the left, we will use the first equation:

$\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)$

By direct substitution:

$=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2$

Right-Hand Limit:

$\displaystyle \lim_{x\to -1^+}f(x)$

Since we are approaching from the right, we will use the second equation:

$=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}$

Direct substitution:

$\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2$

So, we can see that:

$\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2$

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

$\displaystyle \lim_{x \to -1}f(x)=-2$

Our answer is A.

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2 years ago

Find the length of side x in simplest radical from with a rational denominator.​

Find the length of side x in simplest radical from with a rational denominator.