Mister Exam
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How to use it?
- Limit of the function:
-
Limit of (-5+7*n)/(2+n)
-
Limit of (-64+4^x)/(-3+x)
-
Limit of (-7+3*x^2+5*x)/(1+x+3*x^2)
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Limit of (-2+2*x^3+7*x)/(-4-x+3*x^3)
-
Identical expressions
- log(three *x)/x^ two
- logarithm of (3 multiply by x) divide by x squared
- logarithm of (three multiply by x) divide by x to the power of two
- log(3*x)/x2
- log3*x/x2
- log(3*x)/x²
- log(3*x)/x to the power of 2
- log(3x)/x^2
- log(3x)/x2
- log3x/x2
- log3x/x^2
- log(3*x) divide by x^2
Limit of the function log(3*x)/x^2
The solution
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[src]
/log(3*x)\
lim |--------|
x->oo| 2 |
\ x /
$$\lim_{x \to \infty}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right)$$
Limit(log(3*x)/x^2, x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(\log{\left(x \right)} + \log{\left(3 \right)}\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x^{2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(\log{\left(x \right)} + \log{\left(3 \right)}\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x^{2}}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(\log{\left(x \right)} + \log{\left(3 \right)}\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty} x^{2} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(\log{\left(x \right)} + \log{\left(3 \right)}\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x^{2}}\right)$$
=
$$0$$
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
$$\lim_{x \to \infty}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = \log{\left(3 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = \log{\left(3 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = \log{\left(3 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = \log{\left(3 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(3 x \right)}}{x^{2}}\right) = 0$$
More at x→-oo
The graph