1.
We can use L’Hopital’s rule to find the limit of the first expression:
lim x->2 ((x^2 – f(x))/(x – 2)) = lim x->2 (2x – f'(x)) = 2*2 – f'(2) = 4 – f'(2) = 2023
So we have f'(2) = -2019.

Now we can use this information to find the limit of f(x)/x as x approaches 2:
lim x->2 f(x)/x = lim x->2 (f(x) – f(2))/(x – 2) = f'(2) = -2019.

2.
lim x->1 (sqrt(5x^2-4)-1)/(1-sqrt(x))
= lim x->1 (sqrt(5x^2-4)-1)/(1-sqrt(x)) * [sqrt(5x^2-4)+1) / sqrt(5x^2-4)+1)]
= lim x->1 (5x^2-5)/[(1-sqrt(x)) * sqrt(5x^2-4)+1)]
= lim x->1 (5x^2-5)/[(1-sqrt(x)) * sqrt(5x^2-4)+1)] * [(1+sqrt(x)) / (1+sqrt(x))]
= lim x->1 -5(x+1)(1+sqrt(x)) / (sqrt(5x^2-4)+1) => -10

3.
To find A such that g(x) is continuous at x=0, we need to evaluate the limit of g(x) as x approaches 0.

For x < 0, we have x*sin(1/x) which approaches 0 as x approaches 0.

For x > 0, we have x^2*cos(1/x^2) which also approaches 0 as x approaches 0.

Therefore, to make g(x) continuous at x=0, we need to find A such that g(0) = A = 0.

For x < 0, x*sin(1/x) is continuous since both x and sin(1/x) are continuous on (-∞, 0), and their product is continuous on (-∞, 0).

For x > 0, x^2*cos(1/x^2) is also continuous since both x^2 and cos(1/x^2) are continuous on (0, +∞), and their product is continuous on (0, +∞).

Therefore, g(x) is continuous on (-∞, 0) U (0, +∞)

4.
lim x->∞ (5x^2 – x*sin(2x))/(x^2 + 10)
= lim x->∞ (10x – sin(2x) – 2xcos(2x))/2x
= lim x->∞ (10 – 4*cos(2x) + 4xsin(2x)) / 2
= lim x->∞ 5 – sin(2x)/2x – cos(2x) => 5

5.
In this case, we can use the Intermediate Value Theorem to show that the roots of the equation x^3 + 3x^2 – 1 = 0 lie on the interval [-3, 1]. To do this, we need to find two points in the interval such that the value of the function at each of these points has opposite signs.

Let’s start by finding the value of the function at x = -3:
f(-3) = (-3)^3 + 3(-3)^2 – 1 = -1

Next, let’s find the value of the function at x = 1:
f(1) = 1^3 + 3(1)^2 – 1 = 3

Since f(-3) < 0 and f(1) > 0, the Intermediate Value Theorem tells us that there must be at least one root of the equation in the interval [-3, 1].And because when x>1, f(x) is greater than 0, and when x<-3, f(x) is less than 0, so all of the roots of the equation x^3 + 3x^2 – 1 = 0 lie in the interval [-3, 1].

6.
lim x->+∞ |x|/(sqrt(3x^2+1)+5) = |x|/sqrt(3x^2) = 1/sqrt(3)

7.
In order for f(x) to be continuous at x = 0, we need to have:

f(0) exists.
The limit of f(x) as x approaches 0 from the left (x < 0) is equal to f(0).
The limit of f(x) as x approaches 0 from the right (x > 0) is equal to f(0).
Let’s find a, b, and c that satisfy these conditions.

f(0) exists:
lim x->0- a(x+1)^2+1
= a(0+1)^2 + 1
= a + 1

Since we need this limit to be equal to 1+b, we have:

a + 1 = 1 + b

we get:

a = b

The limit of f(x) as x approaches 0 from the right (x > 0) is equal to f(0):
lim x->0+ sin(c/x) = 1+b
This limit exists if and only if c is non-zero. Let’s assume c is non-zero, then we can use L’Hopital’s rule to evaluate the limit:

lim x->0+ sin(c/x) = 0

Therefore, we have:

1+b = 0

so, we get:

a = -1, b = -1, and c can be any value

8.
assume x = 1, then x^2-3x+3 = 1
so lim x->1+ f(x^2-3x+3) = f(1) = 1