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