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-x^2+(2+x)*(3+x)/4

Limit of the function -x^2+(2+x)*(3+x)/4

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      /   2   (2 + x)*(3 + x)\
 lim  |- x  + ---------------|
x->-2+\              4       /
$$\lim_{x \to -2^+}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right)$$
Limit(-x^2 + ((2 + x)*(3 + x))/4, x, -2)
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
One‐sided limits [src]
      /   2   (2 + x)*(3 + x)\
 lim  |- x  + ---------------|
x->-2+\              4       /
$$\lim_{x \to -2^+}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right)$$
-4
$$-4$$
= -4.0
      /   2   (2 + x)*(3 + x)\
 lim  |- x  + ---------------|
x->-2-\              4       /
$$\lim_{x \to -2^-}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right)$$
-4
$$-4$$
= -4.0
= -4.0
Rapid solution [src]
-4
$$-4$$
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to -2^-}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = -4$$
More at x→-2 from the left
$$\lim_{x \to -2^+}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = -4$$
$$\lim_{x \to \infty}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = -\infty$$
More at x→oo
$$\lim_{x \to 0^-}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = \frac{3}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = \frac{3}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = 2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = 2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(- x^{2} + \frac{\left(x + 2\right) \left(x + 3\right)}{4}\right) = -\infty$$
More at x→-oo
Numerical answer [src]
-4.0
-4.0
The graph
Limit of the function -x^2+(2+x)*(3+x)/4