Mister Exam
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How to use it?
- Limit of the function:
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Limit of x^(4/x)
-
Limit of (-6+x^2-x)/(-4+x^2)
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Limit of (-6+x^2-x)/(-21+x+2*x^2)
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Limit of e^(3-x)*(-2+x)
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Identical expressions
- e^(three -x)*(- two +x)
- e to the power of (3 minus x) multiply by ( minus 2 plus x)
- e to the power of (three minus x) multiply by ( minus two plus x)
- e(3-x)*(-2+x)
- e3-x*-2+x
- e^(3-x)(-2+x)
- e(3-x)(-2+x)
- e3-x-2+x
- e^3-x-2+x
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Similar expressions
- e^(3+x)*(-2+x)
- e^(3-x)*(-2-x)
- e^(3-x)*(2+x)
Limit of the function e^(3-x)*(-2+x)
The solution
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/ 3 - x \ lim \E *(-2 + x)/ x->oo
$$\lim_{x \to \infty}\left(e^{3 - x} \left(x - 2\right)\right)$$
Limit(E^(3 - x)*(-2 + x), x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(x - 2\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} e^{x - 3} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(e^{3 - x} \left(x - 2\right)\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\left(x - 2\right) e^{3 - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x - 2\right)}{\frac{d}{d x} e^{x - 3}}\right)$$
=
$$\lim_{x \to \infty}\left(e^{3} e^{- x}\right)$$
=
$$\lim_{x \to \infty}\left(e^{3} e^{- x}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(x - 2\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} e^{x - 3} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(e^{3 - x} \left(x - 2\right)\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to \infty}\left(\left(x - 2\right) e^{3 - x}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x - 2\right)}{\frac{d}{d x} e^{x - 3}}\right)$$
=
$$\lim_{x \to \infty}\left(e^{3} e^{- x}\right)$$
=
$$\lim_{x \to \infty}\left(e^{3} e^{- x}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(e^{3 - x} \left(x - 2\right)\right) = 0$$
$$\lim_{x \to 0^-}\left(e^{3 - x} \left(x - 2\right)\right) = - 2 e^{3}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(e^{3 - x} \left(x - 2\right)\right) = - 2 e^{3}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(e^{3 - x} \left(x - 2\right)\right) = - e^{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(e^{3 - x} \left(x - 2\right)\right) = - e^{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(e^{3 - x} \left(x - 2\right)\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(e^{3 - x} \left(x - 2\right)\right) = - 2 e^{3}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(e^{3 - x} \left(x - 2\right)\right) = - 2 e^{3}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(e^{3 - x} \left(x - 2\right)\right) = - e^{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(e^{3 - x} \left(x - 2\right)\right) = - e^{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(e^{3 - x} \left(x - 2\right)\right) = -\infty$$
More at x→-oo
The graph