Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of (8+x^4+6*x)/(-7+2*x^4+3*x^2)
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Limit of (-1+x^4)/(-1+x+x^4-x^3)
-
Limit of (-1+x^4)/(-1+x^6)
-
Limit of (x^3+4*x)/(x^2+4*x^3)
- Graphing y =:
- 1+x-x^2
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Identical expressions
- one +x-x^ two
- 1 plus x minus x squared
- one plus x minus x to the power of two
- 1+x-x2
- 1+x-x²
- 1+x-x to the power of 2
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Similar expressions
- 1+x+x^2
- 1-x-x^2
Limit of the function 1+x-x^2
The solution
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/ 2\ lim \1 + x - x / x->oo
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right)$$
Limit(1 + x - x^2, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-1 + \frac{1}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-1 + \frac{1}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{u^{2} + u - 1}{u^{2}}\right)$$
=
$$\frac{-1 + 0^{2}}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right) = -\infty$$
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-1 + \frac{1}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-1 + \frac{1}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{u^{2} + u - 1}{u^{2}}\right)$$
=
$$\frac{-1 + 0^{2}}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right) = -\infty$$
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
$$\lim_{x \to \infty}\left(- x^{2} + \left(x + 1\right)\right) = -\infty$$
$$\lim_{x \to 0^-}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(- x^{2} + \left(x + 1\right)\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(- x^{2} + \left(x + 1\right)\right) = 1$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(- x^{2} + \left(x + 1\right)\right) = -\infty$$
More at x→-oo
The graph