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.
Let's go through each of these limits step by step:
1. lim x->0 (x)/(sqrt(x^2+4) - 2)
To evaluate this limit, we can simplify the expression first. Multiply both the numerator and denominator by the conjugate of the denominator, which is sqrt(x^2 + 4) + 2:
lim x->0 (x * (sqrt(x^2 + 4) + 2))/((sqrt(x^2 + 4) - 2) * (sqrt(x^2 + 4) + 2))
After simplification, the expression becomes:
lim x->0 ((x * sqrt(x^2 + 4)) + 2x)/(x^2 + 4 - 4)
Simplifying further, we have:
lim x->0 ((x * sqrt(x^2 + 4)) + 2x)/(x^2)
Now, as x approaches 0, the numerator also approaches 0 since both terms have a factor of x. In the denominator, x^2 approaches 0 as well. Thus, the limit becomes:
lim x->0 ((x * sqrt(x^2 + 4)) + 2x)/(x^2) = 0/0
We can apply L'Hôpital's rule to differentiate the numerator and denominator separately until we obtain a non-indeterminate form. Let's take the derivatives:
lim x->0 ((sqrt(x^2 + 4) + 2) + 2)/(2x) = 4/0
Now, we should distinguish between the right-sided limit (approaching 0 from the positive side, denoted as lim x->0+) and the left-sided limit (approaching 0 from the negative side, denoted as lim x->0-).
For the right-sided limit, lim x->0+, since the numerator is positive while the denominator approaches 0, the fraction goes to positive infinity. So, the answer for the right-sided limit is +∞.
For the left-sided limit, lim x->0-, the numerator is still positive, but the denominator approaching zero will be negative. Hence, the fraction goes to negative infinity. Therefore, the answer for the left-sided limit is -∞.
2. lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2)
Like the first example, let's start by simplifying the expression:
lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2)
By factoring the numerator and denominator, we have:
lim x->1 (x - 3)(x - 2)/(x - 1)(x - 2)
Now, we can cancel out the common factors:
lim x->1 (x - 3)/(x - 1)
By substituting x = 1, we get 0/0, which is an indeterminate form. Applying L'Hôpital's rule, we differentiate the numerator and denominator:
lim x->1 1/(1) = 1
So, the limit as x approaches 1 is 1.
3. lim h->0 [(-7)/(2+h^2) + (7/4)]/h
To evaluate this limit, we can simplify the expression:
lim h->0 [(-7)/(2+h^2) + (7/4)]/h
First, we can simplify the numerator:
[(-7)/(2+h^2) + (7/4)] = [(-7)(4)/(4(2+h^2)) + (7(2+h^2))/(4(2+h^2))]
Simplifying further:
[-28/(8+4h^2) + (14 + 7h^2)/(8+4h^2)] = (-28 + 14 + 7h^2)/(8+4h^2)
Now, let's divide the whole expression by h:
(-28 + 14 + 7h^2)/(8h+4h^3) = (7h^2 - 14)/(4h(h^2+2))
As h approaches 0, the entire numerator becomes 0, and the denominator becomes 0 as well. Therefore, this expression becomes 0/0, which is an indeterminate form.
To evaluate this further, we can apply L'Hôpital's rule. Differentiating the numerator and denominator, we get:
lim h->0 (14h)/(12h^2+8) = 0/8 = 0
Hence, the limit as h approaches 0 is 0.
4. lim x-> neg. infinity (-2x^2 + 3x - 2)/(5x^3 + 4x - x + 1)
To evaluate this limit, we can examine the highest power terms in the numerator and denominator.
Since the highest power term in the numerator is -2x^2, and in the denominator is 5x^3, as x approaches negative infinity, the numerator and denominator both go to positive infinity.
Thus, the limit as x approaches negative infinity is +∞.