Question

Find the indicated limit, if it exists. limit of f of x as x approaches 0 where f of x equals 5 x minus 8 when x is less than 0 and the absolute value of the quantity negative 4 minus x when x is greater than or equal to 0

Answer 1

Find Lim f(x) as x->0

for

f(x)=5x-8 ......x<0

f(x)=|-4-x|......x &ge; 0

Clearly

f(0)=|-4-0|=|-4|=4 exists.

The left-hand limit, as x-> 0- is f(0-)=5(0)-8=-8

The right hand limit, as x->0+ is f(0+)=|-4-0|=|-4|=4

Since the left and right limits do not both equal f(0), the limit does not exist.

Answer 2

The limit does not exists.

Explanation:

Given:

f(x) = 5x - 8 ; x < 0

f(x) = l -4 -x l ; x ≥ 0

Now we have to find the limit when x approaches to 0.

Condition:

If the left hand limit is equal to the right hand limit, then only the limit exists.

Now let's find the left hand limit.

lim 5x - 8

x ---> 0

Now let's apply the limit x = 0, we get

5(0) - 8 = -8

The left hand limit is -8

Now let's find the right hand limit.

lim l -4 - x l

x ----> 0

Now plug in x = 0 in the above absolute function, we get

l -4 - 0 l = l -4 l = 4 [Absolute value of a negative number is positive]

The right hand limit is 4.

Here we can see that the left hand limit does not equal to the right hand limit.

Therefore, the limit does not exists.