The limit as h approaches 0 of (e^(2+h)-e^2)/h is ?

1 answer:

3 0

If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
[e^(2+h) − e^2]/h = [e^2(e^h−1)]/h

so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:

=e^2

You might be interested in

Geometry..already got hw

miss Akunina [59]

the second option

Step-by-step explanation:

Line CD is parallel to line AB, shown by the arrows in the middle of both of the lines.

5 0

3 years ago

49a4 + 87a2b2 + 64b4 TRINOMIO CUADRADO PERFECTO POR ADICION Y SUSTRACCION por favor ayud

KatRina [158]

Answer:

(7a^2 + 8b^2 + 5ab) (7a^2 + 8b^2 - 5ab)

Step-by-step explanation:

Dado que ambos términos son cuadrados perfectos, puede factorizar utilizando la fórmula de la diferencia de cuadrados, a^2 - b^ 2 = (a + b) (a - b), donde a = 7a^2 + 8b^2 y b = 5ab.

English: Since both terms are perfect squares you can factor using the difference of squares formula, a^2 - b^2 = (a + b)(a - b), where a = 7a^2 + 8b^2 and b = 5ab.

6 0

3 years ago

What value ofx is in the solution set of 3(x-4)>5x +2?

kvv77 [185]

$\boldsymbol{\sf{3(x-4) > 5x +2 }}$