Find limit if exists
1. Lim x goes to 0 x^2cos(1/x^3)
2.lim x goes to 1 to the left
4x/x^2+2x-3
3. Lim x goes to π to the right
cscx
#1 cos(1/x^3) oscillates between 1 and -1, but the x^2 out front forces the limit to zero
http://www.wolframalpha.com/input/?i=x^2cos%281%2Fx^3%29+
#2 for x < 1, x^2+2x-3 < 0
So, you have 4*1/(small negative) -> -∞
#3 same thing here. sin(x) > 0 for x<π, so csc(x) -> +∞
You can type in your other functions at wolframalpha to verify the limits.
To find the limits of the given functions, we can use algebraic manipulations and properties of limits.
1. For the limit as x approaches 0 of x^2cos(1/x^3):
First, let's consider the expression inside the cosine function, which is 1/x^3. As x approaches 0, 1/x^3 approaches positive or negative infinity, depending on which side of 0 x is approaching from. This means that the cosine function will oscillate between -1 and 1 infinitely fast as x approaches 0.
Since x^2 is always positive or zero, the product of x^2 and cos(1/x^3) will also oscillate between -x^2 and x^2 infinitely fast. Therefore, the limit does not exist.
2. For the limit as x approaches 1 from the left of 4x/(x^2+2x-3):
To find this limit, we substitute x = 1 into the expression:
lim x → 1⁻ (4x)/(x^2+2x-3) = (4 * 1)/(1^2+2(1)-3) = 4/(1+2-3) = 4/0
Since the denominator becomes zero, the limit is undefined or does not exist.
3. For the limit as x approaches π from the right of cscx:
We can rewrite cscx as 1/sinx. As x approaches π from the right, sinx approaches 0 (sinx becomes positive but very close to zero). This is because sinx oscillates between -1 and 1, and π is a point where sinx crosses the x-axis.
Therefore, the limit can be found by substituting x = π into the expression:
lim x → π⁺ cscx = lim x → π⁺ 1/sinx = 1/sin(π) = 1/0
Since the denominator becomes zero, the limit is undefined or does not exist.
In summary:
1. The limit does not exist.
2. The limit is undefined or does not exist.
3. The limit is undefined or does not exist.