Can anyone check my answer? "f(x) = x4 + 2e^(2x), is continuous at point x = 2...."

[harieche]

Can anyone check my answer? "f(x) = x4 + 2e^(2x), is continuous at point x = 2...."

The question is f(x) = x⁴ + 2e^2x and is continuous at the point x = 2 by using the definition of continuity. You can use, without need to prove, that f(x) is differentiable at x = 2.
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So here what I have done:

f '(x) = lim x→2 [f(x) - f(2)]/(x-2)
then I proved lim x→2 f(x)=f(2)

f(x)- f(a) = [f(x)- f(a)]1 = [f(x)- f(a)](x-2)/(x-2) , for x not = 2

Since f be differentiable at x=2 , we have lim x→2 [f(x) - f(2)]/(x-2) = f '(x). Hence both limits exist. So I apply the propetiy of limit of a product and obtain

lim x→2 f(x) - f(2) = lim x→2 [f(x) - f(2)]/(x-2) lim x→2 (x-2) = f '(2)lim x→2 (x-a) = 0

i.e. lim x→2 f(x) = f(2)

so f (x) is continuous
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I am not sure if i correct or not. Please some one help me if I done something wrong. Thank you.

 

[Ishuda]

The question is f(x) = x⁴ + 2e^2x and is continuous at the point x = 2 by using the definition of continuity. You can use, without need to prove, that f(x) is differentiable at x = 2.
_____________________________________________________
So here what I have done:

f '(x) = lim x→2 [f(x) - f(2)]/(x-2)
then I proved lim x→2 f(x)=f(2)

f(x)- f(a) = [f(x)- f(a)]1 = [f(x)- f(a)](x-2)/(x-2) , for x not = 2

Since f be differentiable at x=2 , we have lim x→2 [f(x) - f(2)]/(x-2) = f '(x). Hence both limits exist. So I apply the propetiy of limit of a product and obtain

lim x→2 f(x) - f(2) = lim x→2 [f(x) - f(2)]/(x-2) lim x→2 (x-2) = f '(2)lim x→2 (x-a) = 0

i.e. lim x→2 f(x) = f(2)

so f (x) is continuous
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I am not sure if i correct or not. Please some one help me if I done something wrong. Thank you.

I would mention in passing that since the f is differential at x=2, f'(2) is bounded. Otherwise, yes a good proof.