finding limit using epsilon and delta of lim x->2 x^3-3x+7=9..plz anyone solve it for me....

In your case, there is no reason you can't calculate the limit of the left side as x->2 by plugging in x = 2 directly. You will find that the value is 9.

If you insist on using the (epsilon, delta) definition of a limit, review the lecture given at:

http://www.khanacademy.org/video/epsilon-delta-limit-definition-2?playlist=Calculus

We solve the inequality: Ix^3-3x+7-9I<e

I(x-2)(x^2+2x+1)I<e
I(x-2)((x-2)^2+6(x-2)+9)I<e
Let e^2/100 +6e/10<1
If Ix-2I<e/10, then
I(x-2)((x-2)^2+6(x-2)+9)I<=
Ix-2I(Ix-2I^2+6Ix-2I+9)< (e/10)(e^2/100+6e/10+9)<(e/10)(1+9)=e,
so we can take delta=epsilon/10

To find the limit using epsilon and delta definition, we can start by rewriting the given limit:

lim x->2 (x^3 - 3x + 7) = 9

Let's introduce ε (epsilon) as the error tolerance and find a Δ (delta) such that if |x - 2| < Δ, then |(x^3 - 3x + 7) - 9| < ε.

Now, let's simplify the expression inside the absolute value:

|(x^3 - 3x + 7) - 9| = |x^3 - 3x - 2|

We want to bound the above expression by ε. Let's factor the expression:

|x^3 - 3x - 2| = |(x - 2)(x^2 + 2x + 1)|

To proceed, we need to make an assumption about Δ or Δ needs to be determined such that |(x - 2)(x^2 + 2x + 1)| < ε.

Since we want x to approach 2, we can assume that |x - 2| < 1. This assumption allows us to simplify the expression further:

|x - 2| < 1
=> -1 < x - 2 < 1
=> 1 < x < 3

Now, let's analyze the quadratic term (x^2 + 2x + 1). Using the fact that |x - 2| < 1 and the range of x we just found, we can determine an upper bound for |x^2 + 2x + 1|:

1 < x < 3
=> 1^2 < x^2 < 3^2
=> 1 < x^2 < 9
=> 1 + 2x + 1 < x^2 + 2x + 1 < 9 + 2x + 1
=> 2 + 2x < x^2 + 2x + 1 < 10 + 2x

Since 2x is smaller than 2x + 1, we can consider the upper bound as 10 + 2x.

Now, we can bound the absolute value expression as:

|x^3 - 3x - 2| = |(x - 2)(x^2 + 2x + 1)|
< |(x - 2)(10 + 2x)|
< |(10 + 2x)(x - 2)|

To get an upper bound for the above expression, we consider the range of x as 1 < x < 3:

1 < x < 3
=> 10 + 2 < 10 + 2x < 10 + 6
=> 12 < 10 + 2x < 16
=> 12(x - 2) < (10 + 2x)(x - 2) < 16(x - 2)

Notice that (10 + 2x) < 0 for (1 < x < 2), so we need to flip the inequality to maintain the direction of the inequality:

12(x - 2) > (10 + 2x)(x - 2) > 16(x - 2)

Now, let's simplify the bounds:

12(x - 2) = 12x - 24
16(x - 2) = 16x - 32

Thus, we have:

12(x - 2) > (10 + 2x)(x - 2) > 16(x - 2)
=> 12x - 24 > (10 + 2x)(x - 2) > 16x - 32

Now, we can manipulate the above expression further:

12x - 24 > (10 + 2x)(x - 2) > 16x - 32
=> 12x - 24 > 10x - 20 > 16x - 32
=> 2x > 4 > x - 8
=> 6 > 2x > -4
=> 3 > x > -2

Now, let's determine an appropriate Δ to make sure that if |x - 2| < Δ, then |(x^3 - 3x + 7) - 9| < ε:

We can choose Δ as the smaller value between 1 and the distance from 2 to the smaller end of the interval we found for x, which is Δ = min(1, 2 - (-2)) = min(1, 4) = 1.

Therefore, if we choose Δ = 1, then whenever |x - 2| < Δ = 1, we have:

|(x^3 - 3x + 7) - 9| = |x^3 - 3x - 2| < |(10 + 2x)(x - 2)| < 12(x - 2) < 12(1) = 12

This means that if |x - 2| < 1, then |(x^3 - 3x + 7) - 9| < 12 for any chosen ε.

So, we have found a Δ = 1 such that whenever |x - 2| < Δ, |(x^3 - 3x + 7) - 9| < ε for any chosen ε. This satisfies the epsilon and delta definition of the limit.

Hence, the limit is given by lim x->2 (x^3 - 3x + 7) = 9.

To find the limit of the function f(x) = x^3 - 3x + 7 as x approaches 2, using the epsilon-delta definition, we need to show that for any given epsilon > 0, there exists a delta > 0 such that whenever 0 < |x - 2| < delta, then |f(x) - 9| < epsilon.

Let's solve this step-by-step:

Step 1: Start with |f(x) - 9| < epsilon

|f(x) - 9| = |(x^3 - 3x + 7) - 9|
= |x^3 - 3x - 2|

Step 2: Factorize the expression

|x^3 - 3x - 2| = |(x - 2)(x^2 + 2x + 1)|

Step 3: Simplify the expression

|x - 2| is always positive, so we can ignore it when considering the absolute value.

|x^3 - 3x - 2| = |(x - 2)(x^2 + 2x + 1)|
= |x^2 + 2x + 1|

Step 4: Find a value for delta

We want |x^2 + 2x + 1| < epsilon.

Since we are looking for a limit as x approaches 2, we can choose a small value for delta such that |x - 2| < delta.

|x^2 + 2x + 1| = |(x + 1)^2|

To ensure |(x + 1)^2| < epsilon, we can choose delta = sqrt(epsilon).

Step 5: Finalize the proof

Now, whenever 0 < |x - 2| < delta = sqrt(epsilon), we have:

|x^2 + 2x + 1| = |(x + 1)^2| < epsilon

Thus, we have proved that the limit of f(x) = x^3 - 3x + 7 as x approaches 2 is 9 using the epsilon-delta definition.