Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of (-1+x)/(x+x^2)
-
Limit of log(-5+x)/log(e^x-e^5)
-
Limit of (-exp(-x)-2*x+exp(x))/(x-sin(x))
- Limit of e^(-n*x)/n
- Integral of d{x}:
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1/(9-x^2)
- Graphing y =:
- 1/(9-x^2)
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Identical expressions
- one /(nine -x^ two)
- 1 divide by (9 minus x squared )
- one divide by (nine minus x to the power of two)
- 1/(9-x2)
- 1/9-x2
- 1/(9-x²)
- 1/(9-x to the power of 2)
- 1/9-x^2
- 1 divide by (9-x^2)
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Similar expressions
- 1/(9+x^2)
Limit of the function 1/(9-x^2)
The solution
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/ 1 \
lim |1*------|
x->-oo| 2|
\ 9 - x /
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right)$$
Limit(1/(9 - x^2), x, -oo)
Detail solution
Let's take the limit
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right)$$ =
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(-1 + \frac{9}{x^{2}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(-1 + \frac{9}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{9 u^{2} - 1}\right)$$
=
$$\frac{0^{2}}{-1 + 9 \cdot 0^{2}} = 0$$
The final answer:
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = 0$$
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right)$$ =
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(-1 + \frac{9}{x^{2}}\right)}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(-1 + \frac{9}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{9 u^{2} - 1}\right)$$
=
$$\frac{0^{2}}{-1 + 9 \cdot 0^{2}} = 0$$
The final answer:
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = 0$$
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
$$\lim_{x \to -\infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = 0$$
$$\lim_{x \to \infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{9}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{9}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{8}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{8}$$
More at x→1 from the right
$$\lim_{x \to \infty}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = 0$$
More at x→oo
$$\lim_{x \to 0^-}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{9}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{9}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{8}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(1 \cdot \frac{1}{9 - x^{2}}\right) = \frac{1}{8}$$
More at x→1 from the right
The graph