What is the limit of (x + SiNx) / x,

Original formula = LIM (1 + SiNx / x)
X tends to infinity
SiNx [- 1,1] oscillation, that is bounded, 1 / X is infinitesimal
So SiNx / X tends to zero
So yuan = 1 + 0 = 1

(x ^ 2-x-2) / (x + 1) x tends to the limit of 2,
X + 1 is up there

lim(x->2)（x^2-x-2)/(x+1)
=lim(x->2)(x+1)(x-2)/(x+1)
=lim(x->2)(x-2)
=0

Limit: LIM (n →∞) (x + 1) (X-2) / (2x + 1) (x-1)

It should be LIM (x →∞) (x + 1) (X-2) / (2x + 1) (x-1)
Divide up and down by X & # 178;
Original formula = LIM (x →∞) (1 + 1 / x) (1-2 / x) / (2 + 1 / x) (1-1 / x)
X tends to zero on all sides of the denominator
So the original formula = (1 + 0) (1-0) / (2 + 0) (1-0) = 1 / 2

Let f (x) = x / [1 + e ^ (1 / x)] and find the limit of F (x) when x tends to 0

When x → 0 +, 1 / X tends to positive infinity, e ^ (1 / x) tends to positive infinity, and f (x) tends to 0
When x → 0 -, 1 / X tends to negative infinity, e ^ (1 / x) tends to 0, limf (x) = 0 / (1 + 0) = 0
So LIM (x → 0) f (x) = 0

What is the left limit of in (1-x) divided by X when x approaches 0?
The piecewise function f (x), when the absolute value of X is less than or equal to 1, f (x) = cos π X / 2, when the absolute value of X is greater than 1, f (x) = the absolute value of X-1
What is the procedure of your second question?

lim ln(1-x)/x
=LIM (- X / x) (LN (1 + T) and T are equivalent infinitesimals, t tends to 0)
=-1
The breakpoint of the second question is x = - 1, which is the jumping breakpoint
If you draw an image, you can see that when x tends to - 1 from the left, f (x) tends to - 2, and when x tends to - 1 from the right, f (x) tends to 0, which is obviously the breakpoint between jumps

Find the value of limit (x tends to 0) LIM (1 / X-1 / (x ^ 2) * in (1 + x))

The original formula = LIM (x - > 0) [x - ln (1 + x)] / X & # 178; lobita rule
= lim(x->0) [1 - 1/(1+x) ] / (2x)
= lim(x->0) (1/2) * 1/(1+x)
= 1/2

The limit of LIM {in (1 + x) ^ 1 / x-ine} / X when x tends to zero

X → 0lim {ln (1 + x) ^ (1 / x) - lne} / x = Lim [ln (1 + x) - x] / x ^ 2 = Lim [ln (1 + x) - ln (e ^ x)] / x ^ 2 = Lim [ln (1 + x) / (e ^ x)] / x ^ 2 = Lim [ln (1 + (1 + x) / (e ^ x) - 1)] / x ^ 2 according to the equivalent infinitesimal: ln (1 + x) ~ x = Lim [(1 + x) / (e ^ x) - 1] / x ^ 2 =

Find the left and right limits when f (x) = x / x, H (x) = | x | / X. and explain whether their limits exist when x tends to 0?
=

F (x) = x / x, left limit = right limit = 1
H (x) = | x | / x, left limit = - 1, right limit = 1, limit does not exist

What is the limit of x times SiNx when x tends to infinity?
Can we say that in the same trend process of X, the result of infinite function and bounded function or infinite function or function whose limit is equal to non-zero constant must be infinite?

The answer is wrong
Limit existence means that when x tends to infinity in any way, the limit value is the same
If xsinx approaches infinity with x = n π, the limit value is 0
When x = 2n π + π / 2 approaches infinity, the limit value is positive infinity
So the limit does not exist

What is the limit of X / SiNx when x tends to infinity

When x goes to infinity, there is no limit
Reason: the value range of SiNx is: [- 1,1]
So there is no limit
In particular, when x tends to zero, the limit is 1

What is the limit of (x + SiNx) / x,