Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of (-16+x^2-6*x)/(-2+x+x^2)
-
Limit of -1+sqrt(5)-x-2/(sqrt(2)-x)
-
Limit of (-cos(x)^3+cos(x))/(4*x*sin(x))
-
Limit of cos((1+x)/x^3)
- Graphing y =:
- 8/(16-x^2)
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Identical expressions
- eight /(sixteen -x^ two)
- 8 divide by (16 minus x squared )
- eight divide by (sixteen minus x to the power of two)
- 8/(16-x2)
- 8/16-x2
- 8/(16-x²)
- 8/(16-x to the power of 2)
- 8/16-x^2
- 8 divide by (16-x^2)
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Similar expressions
- 8/(16+x^2)
Limit of the function 8/(16-x^2)
The solution
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/ 8 \
lim |-------|
x->-oo| 2|
\16 - x /
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)$$
Limit(8/(16 - x^2), x, -oo)
Detail solution
Let's take the limit
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)$$ =
$$\lim_{x \to -\infty}\left(\frac{8 \frac{1}{x^{2}}}{-1 + \frac{16}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{8 \frac{1}{x^{2}}}{-1 + \frac{16}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{8 u^{2}}{16 u^{2} - 1}\right)$$
=
$$\frac{8 \cdot 0^{2}}{-1 + 16 \cdot 0^{2}} = 0$$
The final answer:
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right) = 0$$
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)$$ =
$$\lim_{x \to -\infty}\left(\frac{8 \frac{1}{x^{2}}}{-1 + \frac{16}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{8 \frac{1}{x^{2}}}{-1 + \frac{16}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{8 u^{2}}{16 u^{2} - 1}\right)$$
=
$$\frac{8 \cdot 0^{2}}{-1 + 16 \cdot 0^{2}} = 0$$
The final answer:
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right) = 0$$
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
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right) = 0$$
$$\lim_{x \to \infty}\left(\frac{8}{16 - x^{2}}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{8}{16 - x^{2}}\right) = \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{8}{16 - x^{2}}\right) = \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{8}{16 - x^{2}}\right) = \frac{8}{15}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{8}{16 - x^{2}}\right) = \frac{8}{15}$$
More at x→1 from the right
$$\lim_{x \to \infty}\left(\frac{8}{16 - x^{2}}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{8}{16 - x^{2}}\right) = \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{8}{16 - x^{2}}\right) = \frac{1}{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{8}{16 - x^{2}}\right) = \frac{8}{15}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{8}{16 - x^{2}}\right) = \frac{8}{15}$$
More at x→1 from the right
The graph