Question

Question #42f36

Answer 1

The limit exists and is equal to 0

Explanation:

Put x=3 to find f(3)
f(3)=`3^2-6*3+3`= -6

Given expression is
`(x^2-6x+3 - (-6))/(x-3)`

`(x^2-6x+3 +6)/(x-3)`

`(x^2-6x+9)/(x-3)`

or `(x-3)^2/(x-3)`

Clearly the function isn't defined for x=3 we get 0/0 indeterminate form .

For Right hand limit put x=3+h where h--->0 (h tends to 0)

`(3+h-3)^2/(3+h-3)`
or `h^2/h` [we can easily divide these 'cause h isn't 0 but tends to 0]
or `h`
thus RHL=0 [since h-->0}

Similarly for Left hand limit put x=3-h where h--->0

`(3-h-3)^2/(3-h-3)`
or `h^2/ -h`
or `-h`
thus LHL=0

Since RHL=LHL=0 the limit exists and is equal to 0

Answer 2

`lim_(x->3)(f(x)-f(3))/(x-3)=0`

Explanation:

All right. We have: `lim_(x->3)(f(x)-f(3))/(x-3)`

hmm... let's use another variable, `h` where it equals `x-3`

We have: `lim_(x->3)(f(x)-f(3))/h`

Now, let's use substitution.
We get: `(f(3)-f(3))/0`
Note that `h=x-3`, so when `x=3`, we have `h=3-3=>0`

Since `h` becomes zero when `x=3`, we can say that `f(3)=f(3+h)`

Therefore, we have: `(f(3)-f(3))/0=>(f(3+0)-f(3))/0`

Let's rewrite our original problem.

`lim_(x->3)(f(x)-f(3))/(x-3)=>lim_(x->3)(f(x+h)-f(3))/(h)`

When we plug `3` in the place of `h`, we get:
`(f(3+0)-f(3))/(0)`
hmm... this looks similar to our definition of a derivative:

`lim_(h->0)((f(x+h)-f(x))/h)`

Our limit is there fore asking us, what is the instantaneous rate of change of `f(x)` when `x=3`, or what is `f'(3)`?

Let's find `f'(x)`.

We use the power rule, which states that `d/dx(x^n)=nx^(n-1)` where `n` is a constant.

We have: `f'(x)=2x^(2-1)-6x^(1-1)+3*0x^(0-1)`

=>`f'(x)=2x-6` We can now find `f'(3)`

=>`f'(3)=2*3-6`

=>`f'(3)=0`

That is our answer!