Use definition to prove limx rightarrow 2 x2 3x 6 4Soluti

Use definition to prove lim_x rightarrow 2 (x^2 - 3x + 6) = 4

Solution

Let 0 < |x-2| < , then |f(x) - 4| <

Hence,

|f(x) - 4| <

|x^2 - 3x + 6 - 4| <

|x^2 - 3x + 2| <

|(x-1) (x-2)| <

|x-1| |x-2| <

Now,

|x-2| < / |x-1| { as = |x-2| }

delta can be function of only, not the function of x. So, we will consider when |x 2| < .

Consider some initial value to use as a restriction for like1.

Hence,

< 1.

If < 1,

then,

|x 2| < < 1

|x 2| < 1

1 < x 2 < 1

1 < x < 3.

Now with this domain of values for x, finding largest value of |x1|

Since, upper bound is 2,

|x1| < 2 if x is between 1 and 3. ---- (1)

But as found above the term,

|x-1| | x-2| < -------- (2)

From 1 and 2

Therefore if 2|x2| < then |x1||x2| < because |x1||x2| < 2|x2| <

2 |x-2| <

|x-2| < /2

delta = /2

Main proof :

Given >0,

Let delta = min(/2 , 1)

Consider that if 0 < |x 2| < < 1

then,

1 < x 2 < 1

1 < x < 3

|x 1| < 2.

Now if 0 < |x 2| <

|x 2| < /2

2|x 2| <

|x 1||x 2| < 2|x 1| <

|x1||x2| <

|(x1)(x2)| <

|x ^2 3x+ 2| <

|x^2 3x+ 6 - 4| <

|(x^ 2 3x+ 6)(4)| <

Therefore, lim x2 (x^ 2 3x + 6) = 4