What is the limit of (x ^ 2-1) e ^ (1 / x-1) / X-1 when x tends to 1

What is the limit of (x ^ 2-1) e ^ (1 / x-1) / X-1 when x tends to 1

Happy New Year! 1. The limit of this problem does not exist, d.n.e = do not exist! 2. If the limit exists, both the left and right limits must exist and be equal, & nbsp; & nbsp; & nbsp; and the limit of this problem in x = 1, the left side is 0, and the right side is infinite

When x tends to infinity, find the limit of ((2 + x) e Λ 1 / x) - X

Do change yuan, the solution process is as follows: the key to solve the problem: rewind and change, and then lobita rule. Please adopt if you are satisfied!

When x tends to 1, the limit of F (x) = (x ^ 2-1) / (x-1) * e ^ 1 / (x-1) =?

The calculation is wrong!
f(x)=(x+1)e^[-1/(x-1)]
Limf (x) = infinity when x is 1 (-)
Limf (x) = 0 when x = 1 (+)
So limf (x) doesn't exist

Limx tends to 0 (E - (1 + x) ^ 1 / x) / X to find limit

LIM (x - > 0) (exp (1) - (1 + x) ^ (1 / x)) / x = LIM (x - > 0) (exp (1) - exp (1) exp (LN (1 + x) / x-1)) / x = LIM (x - > 0) exp (1) (1-exp (LN (x + 1) / x-1)) / x = LIM (x - > 0) exp (1) (- (LN (x + 1) / x-1)) / x = LIM (x - > 0) exp (1)

Limx tends to 0 [(a ^ x + B ^ x) / 2] ^ (1 / x), (a > 0, b > 0) to find the limit of the above formula

Limx tends to 0 [(a ^ x + B ^ x) / 2] ^ (1 / x), = limx tends to 0 [1 + (a ^ x + B ^ x-1) / 2-1] ^ (1 / x) = limx tends to 0 [1 + (a ^ x + B ^ x) / 2-1] ^ {1 / [(a ^ x + B ^ x-1) / 2-1]} (1 / x) [(a ^ x + B ^ x-1) / 2-1] base: limx tends to 0 [1 + (a ^ x + B ^ x) / 2-1] ^ {1 / [(a ^ x + B ^ X-1

A limit problem: finding the limit of (e ^ sinx-e ^ x) / (sinx-x) when x → 0~
But how?
"There is a formula, when x → 0, e ^ X-1 ~ x," which is true, but it can only be used in simple multiplication and division operations. It seems that addition and subtraction can not be used so directly.

The second floor method is the most common mistake for beginners: the equivalent infinitesimal can only replace independent factors, that is to say, it can only carry out multiplication and division operations, not addition and subtraction operations. This problem can be solved several times by using the law of Roberta, and the numerator and denominator can be derived respectively. But the most standard method is to use McLaughlin formula to expand each item, as long as it is expanded to the front

Find the limit of (e ^ x-e ^ SiNx) / ((x ^ 2) ln (1 + x)) when x approaches 0

Robida's law is OK. We can find the third derivative. Or we can use Taylor's formula. The result is 1 / 6

(x + e ^ 2x) ^ (1 / SiNx) when x approaches the limit of 0

Take logarithm directly and use the law of Robida; the answer is the third power of E

Finding the limit of Ln (e ^ x + X + 1) tending to 0

Directly substituting x = 0 into in (1 + 0 + 1) = in2

The limit of X tending to 0 + x ^ (1 / ln (e ^ x-1))