Calculus - help with limits

[bmartin2389]

Hey everyone,
So it has been quite awhile since I've taken some math courses but I'm getting into calculus and I was needing some help with a problem I have regarding limits. My main question is
how does the lim 1 as x approaches 0 equal -1?

Here is the full question including work I have already completed.

f(x) = -e^-x if x does not equal 0
1 if x=0 and a=0
find the limit of f(x) from the right and left of the function and graph.

My answers so far:

lim -e^-x = -1
x approaches 0-

The book says for the limit of 1 that the answer is equal to -1, my question is how do they get that answer? I thought if the function was a constant, then the answer to the limit was the constant? or
lim c = c
x approaches a
Maybe its different with piece wise functions? any help is appreciated

 

[Yogi]

As x "approaches" 0, f(x) "approaches" -1. Since we are approaching 0 and never reaching it, the value of f(0)=1 doesn't matter since we never actually make it to x=0. The fact that lim x->0 f(x) is not equal to f(0) just means that the function is discontinuous at x=0.

 

[soroban]

Hello, bmartin2389!

Did you sketch the graph as they requested?

$\displaystyle f(x) \:=\:\begin{Bmatrix} -e^{-x}& \text{if }x \ne 0 \\ 1 & \text{if }x = 0 \end{Bmatrix}$

$\displaystyle \displaystyle\text{Find: }\:\lim_{x\to0}f(x)\,\text{ from the right and left.}$

The graph is that of $\displaystyle y = e^x$ rotated 180^o about the Origin.

                |
                *1
                |
                |
  - - - - - - - + - - - - - - - - -
                |             *
                |     *
                o
            *   |-1
         *      |
       *        |
      *         |
                |
     *          |
                |
                |

The expected y-intercept is (0,-1),
. . but we are told that it is (0,1).
Hence, the function is discontinuous at $\displaystyle x=0.$


Consider $\displaystyle \displaystyle \lim_{x\to0^+} f(x)$
As $\displaystyle x$ approaches 0 from the right, what does $\displaystyle f(x)$ approach?
. . We see that $\displaystyle f(x)$ approaches -1.
As $\displaystyle x$ approaches 0 from the left, what does $\displaystyle f(x)$ approach?
. . We see that $\displaystyle f(x)$ approaches -1.

We can write: .$\displaystyle \displaystyle \lim_{x\to0}f(x) \:=\:-1.$

 

[HallsofIvy]

Hey everyone,
So it has been quite awhile since I've taken some math courses but I'm getting into calculus and I was needing some help with a problem I have regarding limits. My main question is
how does the lim 1 as x approaches 0 equal -1?

Here is the full question including work I have already completed.

f(x) = -e^-x if x does not equal 0
1 if x=0 and a=0
find the limit of f(x) from the right and left of the function and graph.

My answers so far:

lim -e^-x = -1
x approaches 0-

The book says for the limit of 1 that the answer is equal to -1, my question is how do they get that answer? I thought if the function was a constant, then the answer to the limit was the constant? or
lim c = c
x approaches a
Maybe its different with piece wise functions? any help is appreciated

Yes, it is different for functions that are NOT constant! And your function is NOT a constant function.

Recall that the definition of "limit" is "$\displaystyle \lim_{x\to a} f(x)= L$ if and only if, given any $\displaystyle \epsilon> 0$ there exist $\displaystyle \delta> 0$ such that if $\displaystyle 0< |x- a|< \delta$ then $\displaystyle |f(x)- L|< \epsilon$."

Notice the "0< " part (which, unfortunately many people, myself included, don't always write). That excludes the case of x= a so the limit is completely independent of the value of the function at x= a.