Detailed process of LIM (x tends to π / 2) (TaNx / tan3x) finding limit

Detailed process of LIM (x tends to π / 2) (TaNx / tan3x) finding limit

The original two solutions are not concise enough
It's not easy to think of simply using the basic relationship of TaNx = SiNx / cosx with the same angle!
The original formula is immediately changed to: find (SiNx / cosx) / (sin3x / cos3x) (deal with sine first and substitute π / 2)
That is, it is simplified to find the limit of cos3x / cosx when x ∆ π / 2,
Let's use both up and down to find the derivative

Prove Lima ^ (1 / N) = 1 (n tends to infinity) with the definition of limit. Pay attention to prove it with the definition of limit!

An=a^(1/n)
When a = 1, an = 1, take n = 1, for any ε> 0 when n > N, there is always | an-1 | 1
a> The 1 time note an = 1 + HN HN > 0 is expanded by binomial
A = (1 + HN) ^ n = 1 + NHN +... > 1 + NHN
So there is always | an-1 | 1 in HNN
For A1, the above method can also prove an - > 1

Prove Lima ^ n / N with the definition of limit= 0(n→∞) Use the definition (because only about this is learned), that is, let V have e, when n > m, ｜ a ^ n / N! ｜

Since there must be N1, so that when n > = N1, n > |a|, we can only look at the items after N1 (note that when a is given, N1 is a constant)
When n > N1,
|a^n/n!| = | a/1| * .|a/N1| * |a/(N1+1)| * ...|a/n|
<|a|/1 * .|a|/N1 *|a|^(n-N1)/n^(n-N1)
Let | a | / 1 *. | a | / N1 = M
There is a ^ n / N| So Ren Gei ε> 0, take n = N1 + log (|a| / N1) ε/ M (N1 plus a / N1 as the base, ε/ The logarithm of M), so when n > N, there is
|a^n/n!|<= M(|a|/N1)^(N-N1)= ε
So as to obtain evidence

It is proved that when n approaches infinity, the limit of (1 / N) * cos (n π / 2) is 0 By definition

Any E > 0
There is n = [1 / E] + 1, so that when n > N
|(1 / N) * cos (n π / 2) | < = | 1 / n| so when n approaches infinity, the limit of (1 / N) * cos (n π / 2) is 0

Does the arctanx limit exist when x tends to infinity When x tends to infinity and positive infinity, the limit existence of arctanx is equal to π / 2 When x tends to infinity and negative infinity, the limit existence of arctanx is equal to - π / 2 Does this mean that the left and right limits are not equal, so they don't exist? X tends to infinity is divided into two cases: X tends to positive infinity and X tends to negative infinity, but what if the limits of the two cases are not equal? Actually, I'm just obsessed with X tends to positive infinity and X tends to negative infinity Is this similar to the left and right limits?

It is a method to prove the existence of the limit of a function
The left limit is negative infinity and the right limit is positive infinity
You should be a new student and a little confused. Read the concepts in the book again and you will understand

Find the limit that 1 / X * e ^ (1 / x) x tends to 0

X → 0 -, the limit is 0
X → 0 +, the limit is + ∞