consider the function give below
f(x)={ 0, x less then or equal to -2
{ sin 1/x+2 , x> -2
does the lim x->-2+ exist?

Respuesta :

Answer:

The Limit doesn't exist

Step-by-step explanation:

One-Sided Limit

It's the term that describes when the limit of a function is computed as the variable approaches to a value from the right or from the left side.

They are expressed respectively as

[tex]\lim\limits_{x \rightarrow a^{+}}f(x)[/tex]

[tex]\lim\limits_{x \rightarrow a^{-}}f(x)[/tex]

We have a function defined by pieces

f(x)=0 if [tex]x\leq -2[/tex]

f(x)=sin(1/x)+2 if x > -2

For the limit when x approaches to -2 to exist, both one-sided limits must exist and be equal. Let's test the left-side limit

[tex]\lim\limits_{x \rightarrow -2^{-}}f(x)=0[/tex]

Now for the right-side limit

[tex]\lim\limits_{x \rightarrow -2^{+}}sin(1/x)+2=-sin(1/2)+2[/tex]

Since both limits are different, the required limit does not exist