# Calculus Answers To Even Problems

calculus problem help?

how would you do problems like these:
lim (1 + h)^(2/h)
h->0
or

lim[1 + (2/n)]n
n->∞

or even a problem where I have to generalize answers in powers of e like

lim(1 + ah)^(1/h)
h->0

I dont get this at all. plz help

Find the limit of the natural logarithm of the function using la règle de l’Hôpital, and then raise e to the power of that.

lim{h→0} (1 + h)^(2/h)
= lim{h→0} e^ln[(1 + h)^(2/h)]
= e^lim{h→0} ln[(1 + h)^(2/h)]
= e^lim{h→0} (2/h)ln(1 + h)
= e^lim{h→0} 2ln(1 + h) / h
= e^lim{h→0} [2ln(1 + h)]’ / h’
= e^lim{h→0} [2/(1 + h)] / 1
= e^lim{h→0} 2 / (1 + h)
= e^[2 / (1 + 0)]
= e^(2 / 1)
= e²

lim{n→∞} [1 + (2/n)]ⁿ
= lim{n→∞} e^ln[[1 + (2/n)]ⁿ]
= e^lim{n→∞} ln[[1 + (2/n)]ⁿ]
= e^lim{n→∞} n ln[1 + (2/n)]
= e^lim{n→∞} ln[1 + (2/n)] / (1/n)
= e^lim{n→∞} [ln[1 + (2/n)]]’ / (1/n)’
= e^lim{n→∞} [-2/[[n²][1 + (2/n)]]] / (-1/n²)
= e^lim{n→∞} [-2/[1 + (2/n)]] / -1
= e^lim{n→∞} 2 / [1 + (2/n)]
= e^[2 / [1 + (2/∞)]]
= e^[2 / (1 + 0)]
= e^(2 / 1)
= e²

lim{h→0} (1 + ah)^(1/h)
= lim{h→0} e^ln[(1 + ah)^(1/h)]
= e^lim{h→0} ln[(1 + ah)^(1/h)]
= e^lim{h→0} (1/h)ln(1 + ah)
= e^lim{h→0} ln(1 + ah) / h
= e^lim{h→0} [ln(1 + ah)]’ / h’
= e^lim{h→0} [a/(1 + ah)] / 1
= e^lim{h→0} a / (1 + ah)
= e^[a / (1 + a(0))]
= e^[a / (1 + 0)]
= e^(a / 1)
= e^a

Calculus II- Integrate by Trigonometric Substitution