Question

Can someone give me the answers to problems a-l? I already finished it but I don't have an answer key so I want to check with someone. Thanks.

Answer 1

If we approach 1 from the left, we're using the blue function, but if we approach 1 from the right, we're using the green function. So, we have

$\displaystyle \lim_{x\to 1^-}f(x) = 0,\quad\lim_{x\to 1^+}f(x) = -1$

Since the left and right limits are different, the limit

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

does not exist.

When we approach 0, we always use the blue function. Both halves of the blue function tend to 1 as x approaches zero. So, in this case, we have

$\displaystyle \lim_{x\to 0^-}f(x) = \lim_{x\to 0^+}f(x) = 1 = \lim_{x\to 0}f(x)$

Note that the fact that, by definition, we have $f(0)=2$ doesn't mean that the limit is wrong, or that it doesn't exist. It simply means that the function is not continuous, because we have

$\displaystyle f(x_0)\neq \lim_{x\to x_0}f(x)$

As for -2, x can approach this value only from the left, because the function is not defined between -2 and -1. So, we have

$\displaystyle \lim_{x\to -2}f(x) = \lim_{x\to -2^-}f(x) 0-\infty$

The limit as x approaches 3 is similar to the one where x approaches zero: the function is not defined at x=3, but the limit from both sides approaches -1:

$\displaystyle \lim_{x\to 3^-}f(x) = \lim_{x\to 3^+}f(x) = -1 = \lim_{x\to 3}f(x)$

As for the limits as x approaches infinity, we have to deduce from the graph that the funtion grows indefinitely as x grows, i.e.

$\displaystyle \lim_{x\to +\infty}f(x) = +\infty$

And that the function has no limit as $x\to-\infty$, because it has a sinusoidal behaviour, with an ever-growing amplitude.