To the - x power of 3, X → + infinity. Does it tend to infinity or infinity

To the - x power of 3, X → + infinity. Does it tend to infinity or infinity

The negative x power of three is equal to one third of the x power of three. Infinity is close to zero, but greater than zero

Why is the 1 / x power of e neither infinity nor infinitesimal when x tends to zero?

When x tends from positive to 0, then 1 / X tends to positive infinity, and the 1 / x power of e tends to positive infinity;
When x tends from negative to 0, then 1 / x approaches negative infinity, and the 1 / x power of e tends to 0

Why is limx → 1 x ^ 2 / 1-x not equal to 1? I think when x approaches 1, 1-x is infinitesimal and 2 / 1-x is infinite, so the infinite power of 1 is 1,

There are several limits that need specific analysis and cannot be generalized. These conditions can be abbreviated as (the value after the limit is taken for each part) ∞ / ∞, ∞ × 0,0 / 0,1 ^ ∞, ∞ ^ 1. Because these cases show that there are ∞ or infinitesimal terms of the same order that can be eliminated from each other. These types can finally be transformed into the form of ∞ / ∞ or 0 / 0 (exponential or power class is transformed into the form of ∞ / ∞ or 0 / 0 through logarithm), then the common factor is eliminated in ∞ / ∞ or 0 / 0, and then it can be calculated. It must not be simply considered that the infinity of 1 is 1

Draw the graph of function f (x) = {2x + 1 x ≥ 0 2x-1 x < 0, and judge whether the limit of the function exists when x → 0?

When   If x ≥ 0, draw a graph with y = 2x + 1 and keep it   X ≥ 0;      When   x＜0   Draw a graph with y = 2x-1 and keep it   x＜0   part;
The image of the above two parts is the image of the function
When x ≥ 0, the limit of X → 0 is 1; When x < 0, the limit of X → 0 is - 1;
Therefore, the limit of the function does not exist when x → 0

Piecewise function, when x

Because LIM (x - > 0) f (x) exists
So LIM (x - > 0 -) f (x) = LIM (x - > 0 +) f (x)
lim(x->0-)f(x)=lim(x->0-)e^x=1
lim(x->0+)f(x)=lim(x->0+)2x+a=a
So a = 1

Let the piecewise function f (x) = the square of X + A, x > = 1- 2x-1,x

When x ≥ 1, f (x) = x ²+ a
When x