Mister Exam
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How to use it?
- Limit of the function:
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Limit of -2+x^3+6*x
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Limit of ((-2+x)/x)^(2*x)
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Limit of -6+x^2+5*x
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Limit of (2+x^2-3*x)/(4+x^2-5*x)
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Identical expressions
- four - six *x
- 4 minus 6 multiply by x
- four minus six multiply by x
- 4-6x
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Similar expressions
- 4+6*x
Limit of the function 4-6*x
The solution
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lim (4 - 6*x) x->oo
$$\lim_{x \to \infty}\left(4 - 6 x\right)$$
Limit(4 - 6*x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(4 - 6 x\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(4 - 6 x\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-6 + \frac{4}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-6 + \frac{4}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{4 u - 6}{u}\right)$$
=
$$\frac{-6 + 0 \cdot 4}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(4 - 6 x\right) = -\infty$$
$$\lim_{x \to \infty}\left(4 - 6 x\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(4 - 6 x\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-6 + \frac{4}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-6 + \frac{4}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{4 u - 6}{u}\right)$$
=
$$\frac{-6 + 0 \cdot 4}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(4 - 6 x\right) = -\infty$$
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
$$\lim_{x \to \infty}\left(4 - 6 x\right) = -\infty$$
$$\lim_{x \to 0^-}\left(4 - 6 x\right) = 4$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(4 - 6 x\right) = 4$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(4 - 6 x\right) = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(4 - 6 x\right) = -2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(4 - 6 x\right) = \infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(4 - 6 x\right) = 4$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(4 - 6 x\right) = 4$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(4 - 6 x\right) = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(4 - 6 x\right) = -2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(4 - 6 x\right) = \infty$$
More at x→-oo
The graph