Let f(x) = \left\{\begin{matrix} \frac {x+2}{x+2}, if < -2\\ 2x, if -2 \leq x < 3 \\ 2x-2, if x>3...

Question:

Let {eq}f(x) = \left\{\begin{matrix} \frac {x+2}{x+2}, \ if \ < -2\\ 2x, \ if \ -2 \leq x < 3 \\ 2x-2, \ if \ x>3 \end{matrix} \right. {/eq}

a) Find {eq}\lim f(x) {/eq} as {eq}x {/eq} approaches {eq}-2 {/eq} from left.

b) Does {eq}\lim f(x) {/eq} as {eq}x {/eq} approaches {eq}-2 {/eq} exist? Why or why not.

c) Does {eq}\lim f(x) {/eq} as {eq}x {/eq} approaches {eq}3 {/eq} exist? Why or why not.

d) Is {eq}f(x) {/eq} continuous at {eq}x=-2 {/eq}? Why or why not?

Limits of a Function:

The limit in one of the calculus ' very important concepts is used to find the value of the function's graph at the point where it is broken or discontinuous. So there are limits if we can find the function's approaching value.

Answer and Explanation:

a) The left hand limid of the given function is {eq}\lim_{x\rightarrow -2 ^-} f(x)=\infty \\ {/eq}

b) The right hand limit is {eq}\lim_{x\rightarrow -2 ^+} f(x)=-4 \\ {/eq}

{eq}\lim_{x\rightarrow -2 ^-} f(x) \neq \lim_{x\rightarrow -2 ^+} f(x)\\ \lim_{x\rightarrow -2} f(x)=DNE {/eq}

c) {eq}\lim_{x\rightarrow 3 ^-} f(x)=6\\ \lim_{x\rightarrow 3 ^+} f(x)=4\\ \lim_{x\rightarrow 3 ^-} f(x) \neq \lim_{x\rightarrow 3 ^+} f(x)\\ {/eq}

The left hand limit not equal to the right hand limit. So the limit does not exist at {eq}x=3 {/eq}

d) The given function is not continuous at {eq}x=-2 {/eq}

Because, the limit of the function at {eq}x=-2 {/eq} does not exist.


Learn more about this topic:

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How to Determine if a Limit Does Not Exist

from AP Calculus AB: Exam Prep

Chapter 4 / Lesson 9
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