Are these correct?

lim x->0 (x)/(sqrt(x^2+4) - 2)
I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity?

lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2)
I get 2/0, so lim x-> 1+ = - infinity? and lim x->1- = + infinity?

lim h->0 [(-7)/(2+h^2) + (7/4)]/h
I used a computational website to get (7/4) as the answer, but I did not get this. My work ends with: (-28)/(4(h+8) + (7(h+8)/(4(h+8)) and I end with -5.25 for an answer?

Last one!

lim x-> neg. infinity (-2x^2 + 3x - 2)/(5x^3 + 4x -x + 1)
Don't even know about this one?

Please help. I would like to understand these.

The first two are correct.

Looking at the pattern of the third and considering the answer , I think you have a typo
The function appears to be f(x) = -7/x^2 and you are finding the derivative when x = 2

Instead of lim h->0 [(-7)/(2+h^2) + (7/4)]/h , it should be
lim h->0 [(-7)/(2+h)^2 + (7/4)]/h , notice the change in brackets.

then I get [-28 + 7(2+h)^2]/(4(2+h)^2) / h
= [ -28 + 28 + 28h + 7h^2]/(4(2+h)^2) / h
= h(28+7h)/(4(2+h)^2) / h
= (28+7h)/(4(2+h)^2
now as h ---> 0 this becomes
28/16 = 7/4

for the last one, how about dividing each term in both the numerator and denominator by the highest power of x that you see, that is , by x^3
to get the expression as

(-2/x + 3/x^2 - 2/x^3)/(5 + 4/x^2 - 1/x^2 + 1/x^3)

Now consider each term
As x becomes hugely negative, each of the terms would still be negative, but very very close to zero, so the top approaches zero, the bottom obvious approaches 5
so you have -0/5 which approaches 0 from the negative.

Let's go through each of these problems and determine the correct limit.

1. lim x→0 (x)/(sqrt(x^2+4) - 2)

To find this limit, we can multiply the numerator and denominator by the conjugate of the denominator, which is √(x^2+4) + 2:

lim x→0 (x)/(sqrt(x^2+4) - 2) * (sqrt(x^2+4) + 2)/(sqrt(x^2+4)+2)

Simplifying the expression gives:

lim x→0 (x * (sqrt(x^2+4) + 2))/((x^2+4)-4)

lim x→0 (x * (sqrt(x^2+4) + 2))/(x^2)

Now, plug in x = 0 into the expression to find the limit:

lim x→0 (0 * (sqrt(0^2+4) + 2))/(0^2)

lim x→0 (0 * (2))/(0)

lim x→0 0

The limit is 0.

2. lim x→1 (x^2 - 5x + 6)/(x^2 - 3x + 2)

Similarly, we can factorize the numerator and denominator:

lim x→1 ((x-2)(x-3))/((x-1)(x-2))

Canceling out the common factor (x-2) gives:

lim x→1 (x-3)/(x-1)

At this point, we can directly substitute x = 1 into the expression:

lim x→1 (1-3)/(1-1)

lim x→1 -2/0

The denominator is 0, indicating that the limit is approaching infinity or negative infinity. We need to examine the left-hand limit (x→1-) and right-hand limit (x→1+).

For x→1-, we can use values slightly less than 1:

lim x→1- (x-3)/(x-1) = (-)/(-) = +infinity

For x→1+, we can use values slightly greater than 1:

lim x→1+ (x-3)/(x-1) = (+)/(-) = -infinity

So the left-hand limit approaches positive infinity, and the right-hand limit approaches negative infinity.

3. lim h→0 [(-7)/(2+h^2) + (7/4)]/h

Let's simplify this expression step by step. First, simplify the numerator:

[-7/(2+h^2) + (7/4)]

To add the two fractions, we need to have a common denominator:

[(-7*4)/(4(2+h^2)) + (7(2+h^2))/(4(2+h^2))]

Simplifying the numerator gives:

[(-28)/(8+4h^2) + (14+7h^2)/(4(2+h^2))]

Now, simplify further by finding the common denominator for the fraction:

[(-28)/(8+4h^2) + (14+7h^2)/(4+2h^2)]

Combining the two fractions gives:

[(-28+14+7h^2)/(8+4h^2)]

Simplifying the numerator:

[-14+7h^2/(8+4h^2)]

Finally, divide the entire expression by h:

[(-14+7h^2)/(h(8+4h^2))]

Now, we can find the limit by plugging in h = 0:

(-14+7(0^2))/(0(8+4(0^2)))

(-14+7(0))/(0(8+4(0)))

(-14+0)/(0(8+0))

-14/0

The denominator is 0, indicating that the limit is approaching infinity or negative infinity. Let's analyze the left-hand limit (h→0-) and the right-hand limit (h→0+).

For h→0-, use values slightly less than 0:

(-14+7(0))/(0(8+4(0))) = -14/0 = -infinity

For h→0+, use values slightly greater than 0:

(-14+7(0))/(0(8+4(0))) = -14/0 = -infinity

Therefore, both the left-hand limit and the right-hand limit approach negative infinity.

4. lim x→-∞ (-2x^2 + 3x - 2)/(5x^3 + 4x - x + 1)

To determine this limit, we check the highest power of x in the numerator and the denominator, which is x^2 and x^3, respectively.

As x approaches negative infinity, the denominator, which has a higher power of x, will dominate the expression. Hence, the limit will behave similar to the limit of the term with the highest power term in the denominator.

So, as x→-∞, the term (5x^3) will have a greater effect, and we can consider only this term:

lim x→-∞ (5x^3 + 4x - x + 1)

Now, we can drop the lower-order terms and find the limit:

lim x→-∞ (5x^3)

As x→-∞, the term (5x^3) approaches negative infinity. Therefore, the overall limit as x approaches negative infinity will also be negative infinity.

I hope this explanation helps you understand how to determine limits in calculus. If you have any further questions, please feel free to ask!