[] calculate the following two limits 1. Lim [(2x + 3) / (2x + 1)] ^ (x + 1) x approaches infinity 2. Lim{[e ^ (2x) + e ^ (- 2x) - 2] / [1-cosx]} x approaches 0 It's best not to insert screenshots, there are audits

[] calculate the following two limits 1. Lim [(2x + 3) / (2x + 1)] ^ (x + 1) x approaches infinity 2. Lim{[e ^ (2x) + e ^ (- 2x) - 2] / [1-cosx]} x approaches 0 It's best not to insert screenshots, there are audits

The first uses the special limit and the second uses the Robita rule
(x→∞) lim[(2x+3)/(2x+1)]^(x+1)
=lim[1+1/(x+1/2)]^(x+1)
=lim([1+1/(x+1/2)]^(x+1/2)*[1+1/(x+1/2)]^(1/2))
=e
(x→0) lim{[e^(2x)+e^(-2x)-2]/[1-cosx]}
=lim{[2e^(2x)-2e^(-2x)]/sinx}
=lim{4e^(2x)+4e(-2x)]/cosx}
=(4+4)/1
=8

Ask a question to calculate the high number of the limit LIM (1 + 1 / 2 + 1 / 4 +... + 1 / 2n -) n tends to infinity, and N - represents the nth power of 2,

Summation of proportional series
Original formula = LIM ((1 - (1 / 2) ^ n) / (1-1 / 2)) = 1 / (1 / 2) = 2

Help solve the following limit calculations lim（ln Sin x / ln x ) X → 0 x tends to 0

lim(x→0)（ln Sin x / ln x ) =lim(x→0)（cosx/sinx)/(1/x)=1

How to prove that this limit does not exist? lim(x->+∞)√[(x^2+1)/x]=

[(x ^ 2 + 1) / x] = x + 1 / X ≥ 2, the equal sign holds only when x = 1
Therefore, LIM (x - > + ∞) √ [(x ^ 2 + 1) / x] = LIM (x - > + ∞) √ (x + 1 / x) = LIM (x - > + ∞) √ x = + ∞ does not exist

Prove whether this simple limit exists? LIM (n tends to positive infinity) BN = 1 LIM (n tends to positive infinity) CN = 0 LIM (n tends to positive infinity) bncn does not exist, Why? Please explain it with the method of teaching idiots + your knowledge,

If limbn and limcn exist, LIM (n → + ∞) bncn = [LIM (n → + ∞)] [bnlim (n → + ∞) BN] = 1 × 0 = 0, otherwise. That is, when LIM (n → + ∞) bncn exists, limbn and limcn do not necessarily exist. For example, BN = n, CN = 1 / n ², Then LIM (n → + ∞) bncnl

Judgment limit does not exist 1 / (1 + 2 ^ (1 / x)) how to judge that his limit does not exist? Um... Sorry, X tends to zero

The limit is 0 when X - > 0 +
The limit is 1 when X - > 0 -
So when X - > 0, the function limit does not exist