Question

It's 11 that grade, please help me I'm stuck ​

Answer 1

Answer:

Limit $\lim_{x \to 0} f(x)$ does not exist.

Step-by-step explanation:

To calculate left hand limit, we use a value slightly lesser than that of 0.

To calculate right hand limit, we use a value slightly greater than that of 0.

Let h be a very small value.

Left hand limit will be calculate at 0-h

Right hand limit will be calculate at 0+h

First of all, let us have a look at the value of f(0-h) and f(0+h)

$f(0-h)=f(-h) = \dfrac{-h}{|-h|}\\\Rightarrow \dfrac{-h}{h} = -1$

$f(0-h)=-1 ....... (1)$

$f(0+h)=f(h) = \dfrac{h}{|h|}\\\Rightarrow \dfrac{h}{h} = 1$

$f(0+h)=1 ....... (2)$

Now, left hand limit:

$\lim_{x \to 0^{-} } f(x)$ = $\lim_{h \to 0} f(0-h)$

$\Rightarrow$ $\lim_{h \to 0} f(-h)$

Using equation (1):

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

Now, Right hand limit:

$\lim_{x \to 0^{+} } f(x)$ = $\lim_{h \to 0} f(0+h)$

$\Rightarrow$ $\lim_{h \to 0} f(h)$

Using equation (2):

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

Since Left Hand Limit $\neq$ Right Hand Limit

So, the answer is:

Limit $\lim_{x \to 0} f(x)$ does not exist.