# calculus

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

1. 👍
2. 👎
3. 👁
1. 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:

1. 👍
2. 👎
2. 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

1. 👍
2. 👎

## Similar Questions

1. ### Quick calc question

The Riemann sum, the limit as the maximum of delta x sub i goes to infinity of the summation from i equals 1 to n of f of the quantity x star sub i times delta x sub i , is equivalent to the limit as n goes to infinity of the

2. ### Calculus

Let f be a function defined for all real numbers. Which of the following statements must be true about f? Which might be true? Which must be false? Justify your answers. (a) lim of f(x) as x approaches a = f(a) (b) If the lim of

1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 ***The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 ***2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 ***9 The

4. ### AP Calculus AB Part 2

The Riemann sum, the limit as the maximum of delta x sub i goes to infinity of the summation from i equals 1 to n of f of the quantity x star sub i times delta x sub i , is equivalent to the limit as n goes to infinity of the

1. ### Calculus Limits

Question: If lim(f(x)/x)=-5 as x approaches 0, then lim(x^2(f(-1/x^2))) as x approaches infinity is equal to (a) 5 (b) -5 (c) -infinity (d) 1/5 (e) none of these The answer key says (a) 5. So this is what I know: Since

2. ### Check my CALCULUS work, please! :)

Question 1. lim h->0(sqrt 49+h-7)/h = 14 1/14*** 0 7 -1/7 Question 2. lim x->infinity(12+x-3x^2)/(x^2-4)= -3*** -2 0 2 3 Question 3. lim x->infinity (5x^3+x^7)/(e^x)= infinity*** 0 -1 3 Question 4. Given that: x 6.8 6.9 6.99 7.01

3. ### Calculus

lim (arccsinx)(cotx) x-->0+ How do I solve this limit using L'Hopital's rule?

4. ### Calculus

Consider the function f(x)=(5^x−1)/x. A) Fill in the table values for f(x): x= -0.1, -0.01, -0.001, -0.0001, 0.0001, 0.001, 0.01, 0.1 f(x)= 1.4866, 1.5866, 1.6081, 1.6093, 1.6096, 1.6107, 1.6225, 1.7462 B) Based on the table

1. ### Calculus

Find the limit. lim 5-x/(x^2-25) x-->5 Here is the work I have so far: lim 5-x/(x^2-25) = lim 5-x/(x-5)(x+5) x-->5 x-->5 lim (1/x+5) = lim 1/10 x-->5 x-->5 I just wanted to double check with someone and see if the answer is

2. ### calc 1

For the limit lim x → 2 (x3 − 5x + 3) = 1 illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.) i set it up like this