Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of ((2+x)/(4+x))^cos(x)
-
Limit of -5-2*x^2+8*x
-
Limit of (-9+x^2)/(-27+x^3)
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Limit of 1+7*x+11*x^2/2
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Identical expressions
- two ^(-x)*log(two *x)
- 2 to the power of ( minus x) multiply by logarithm of (2 multiply by x)
- two to the power of ( minus x) multiply by logarithm of (two multiply by x)
- 2(-x)*log(2*x)
- 2-x*log2*x
- 2^(-x)log(2x)
- 2(-x)log(2x)
- 2-xlog2x
- 2^-xlog2x
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Similar expressions
- 2^(x)*log(2*x)
Limit of the function 2^(-x)*log(2*x)
The solution
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/ -x \ lim \2 *log(2*x)/ x->oo
$$\lim_{x \to \infty}\left(2^{- x} \log{\left(2 x \right)}\right)$$
Limit(log(2*x)/2^x, x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} \log{\left(2 x \right)} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} 2^{x} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(2^{- x} \log{\left(2 x \right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(2 x \right)}}{\frac{d}{d x} 2^{x}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{2^{- x}}{x \log{\left(2 \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{2^{- x}}{x \log{\left(2 \right)}}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty} \log{\left(2 x \right)} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} 2^{x} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(2^{- x} \log{\left(2 x \right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(2 x \right)}}{\frac{d}{d x} 2^{x}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{2^{- x}}{x \log{\left(2 \right)}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{2^{- x}}{x \log{\left(2 \right)}}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(2^{- x} \log{\left(2 x \right)}\right) = 0$$
$$\lim_{x \to 0^-}\left(2^{- x} \log{\left(2 x \right)}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(2^{- x} \log{\left(2 x \right)}\right) = -\infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(2^{- x} \log{\left(2 x \right)}\right) = \frac{\log{\left(2 \right)}}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(2^{- x} \log{\left(2 x \right)}\right) = \frac{\log{\left(2 \right)}}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(2^{- x} \log{\left(2 x \right)}\right) = \infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(2^{- x} \log{\left(2 x \right)}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(2^{- x} \log{\left(2 x \right)}\right) = -\infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(2^{- x} \log{\left(2 x \right)}\right) = \frac{\log{\left(2 \right)}}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(2^{- x} \log{\left(2 x \right)}\right) = \frac{\log{\left(2 \right)}}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(2^{- x} \log{\left(2 x \right)}\right) = \infty$$
More at x→-oo
The graph