Mister Exam
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- Limit of the function:
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Limit of ((5+x^2-6*x)/(5+x^2-5*x))^(2+3*x)
-
Limit of (2+x^2+3*x)/(-4+x^2)
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Limit of (-8+x^2+2*x)/(-2+x)
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Limit of (1+x)^(2/3)-(-1+x)^(2/3)
- 1-3*x
-
Identical expressions
- one - three *x
- 1 minus 3 multiply by x
- one minus three multiply by x
- 1-3x
-
Similar expressions
- 1+3*x
Limit of the function 1-3*x
The solution
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lim (1 - 3*x) x->oo
$$\lim_{x \to \infty}\left(1 - 3 x\right)$$
Limit(1 - 3*x, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(1 - 3 x\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(1 - 3 x\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-3 + \frac{1}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-3 + \frac{1}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{u - 3}{u}\right)$$
=
$$\frac{-3}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(1 - 3 x\right) = -\infty$$
$$\lim_{x \to \infty}\left(1 - 3 x\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(1 - 3 x\right)$$ =
$$\lim_{x \to \infty}\left(\frac{-3 + \frac{1}{x}}{\frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{-3 + \frac{1}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{u - 3}{u}\right)$$
=
$$\frac{-3}{0} = -\infty$$
The final answer:
$$\lim_{x \to \infty}\left(1 - 3 x\right) = -\infty$$
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
$$\lim_{x \to \infty}\left(1 - 3 x\right) = -\infty$$
$$\lim_{x \to 0^-}\left(1 - 3 x\right) = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(1 - 3 x\right) = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(1 - 3 x\right) = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(1 - 3 x\right) = -2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(1 - 3 x\right) = \infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(1 - 3 x\right) = 1$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(1 - 3 x\right) = 1$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(1 - 3 x\right) = -2$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(1 - 3 x\right) = -2$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(1 - 3 x\right) = \infty$$
More at x→-oo
The graph