Can anyone evaluate these two calculus problem?

Question

lim (2 - x)
x =>2+ |2 - x|


and


lim (e^3x - 1)/(e^x - 1)
x=>0


I got 2
for the first one
and
1/3 for the second

Answer 1

Let me try the second problem first, as a L'Hospital case

taking the derivative of the numerator and denominator separately

derivative of numerator...d/dx(e^(3x)-1) = 3e^3x
derivative of the denominator ...d/dx(e^x -1) = e^x

and finding the limit as x goes to zero
and the limit is 3

For the first problem, I will work it this way

lim(2-x) as x goes to 2 + abs(2-x)

limit (2-2-abs(2-x))

lim (abs(2-x)) = abs(lim(2)) - abs(lim(x))

= 2 - abs(lim(x))

Good luck.

Answer 2

The formatting of the expression in your first limit is screwed up. I can't tell what you are asking.

For the second expression, use l'Hopital's rule:

lim x->0 of (exp(3x)-1)/(exp(x)-1)
= lim x-> 0 of (3*exp(3x))/exp(x)
= 3*1/1 = 3

Answer 3

#1, i get 2 also

#2

[e^3x-1]/[e^x-1]

e^3x/e^x
e^2x, as x->0, limit approaches 1

Answer 4

combine via factors: u = x . . . . . . dv = f''(x) du = dx . . . . v = f'(x) ??¹ x f''(x) dx = x f'(x) |?¹ ? ??¹ f'(x) dx . . . . . . . . . . = (a million*f'(a million) ? 0*f'(0)) ? f(x) |?¹ . . . . . . . . . . = f'(a million) ? (f(a million) ? f(0)) . . . . . . . . . . = 2 ? (5 ? 6) . . . . . . . . . . = 3