There are several ways to find the limit of calculus in freshman year

There are several ways to find the limit of calculus in freshman year

1. Definition method
2. When the numerator denominator tends to zero or infinity, we use lobta's law to calculate the derivative of the numerator denominator at the same time
3. Pinch theorem
4. Equivalent infinitesimal. This is the most used problem http://baike.baidu.com/view/2003648.htm
5. The method of dividing the numerator and denominator by one X ^ n is not available on the Internet
6. In this case, limf (x) x --- x0 = f (x0)
7. If it is similar to radical (f (x)) + radical (g (x)), it is better to use molecular rationalization
The above method, you copy to Baidu search, very detailed
The above is my own common method, the answer is not good

Limit proof of calculus
Prove that the limit of sin1 / N to infinity is 0 (by definition)
Sorry, there's one more question.
It is proved that the limit of LNX tending to 1 is 0

|sin n|≤1
So | Sinn / N | ≤ | 1 / N | = 1 / n
Take any small positive number ε
If 1 / N = ε, then n = 1 / ε
When n > N, 1 / N is obtained

Limit of calculus
LIM (x tends to infinity) (x2 + (cosx) 2-1) / (x + SiNx) 2=

LIM (x - > OO) (x ^ 2 + (cosx) ^ 2-1) / (x + SiNx) ^ 2 = LIM (x - > OO) (x ^ 2 - (SiNx) ^ 2) / (x + SiNx) ^ 2 numerator denominator is divided by xsinx to get: = Lim (x - > OO) (x / SiNx SiNx / x) / (x / SiNx + SiNx / x + 2) = LIM (x - > OO) (x / SiNx + SiNx / x + 2-2-2sinx / x) / (x / SiNx + sin)