Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of 1-cos(3*x)
-
Limit of ((6+5*x)/(-1+5*x))^((1+2*x^2)/x)
-
Limit of (-3*5^(1+x)+5*2^x)/(2*5^x+100*2^x)
-
Limit of (4^x-9^x)/(4^(1+x)+9^(1+x))
- Integral of d{x}:
- log(1+x)/x^2
- Graphing y =:
- log(1+x)/x^2
-
Identical expressions
- log(one +x)/x^ two
- logarithm of (1 plus x) divide by x squared
- logarithm of (one plus x) divide by x to the power of two
- log(1+x)/x2
- log1+x/x2
- log(1+x)/x²
- log(1+x)/x to the power of 2
- log1+x/x^2
- log(1+x) divide by x^2
-
Similar expressions
- log(1-x)/x^2
- (-x*e^x+log(1+x))/x^2
- (x^2-log(1+x))/x^2
- (-x+log(1+x))/x^2
Limit of the function log(1+x)/x^2
The solution
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[src]
/log(1 + x)\
lim |----------|
x->oo| 2 |
\ x /
$$\lim_{x \to \infty}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right)$$
Limit(log(1 + x)/x^2, x, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} \log{\left(x + 1 \right)} = \infty$$
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.
$$\lim_{x \to \infty}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x + 1 \right)}}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x \left(x + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{x + 1}}{\frac{d}{d x} 2 x}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{2 \left(x^{2} + 2 x + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{2 \left(x^{2} + 2 x + 1\right)}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{x \to \infty} \log{\left(x + 1 \right)} = \infty$$
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.
$$\lim_{x \to \infty}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x + 1 \right)}}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{2 x \left(x + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{x + 1}}{\frac{d}{d x} 2 x}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{2 \left(x^{2} + 2 x + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(- \frac{1}{2 \left(x^{2} + 2 x + 1\right)}\right)$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
The graph
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = 0$$
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = \log{\left(2 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = \log{\left(2 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = 0$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = -\infty$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = \infty$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = \log{\left(2 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = \log{\left(2 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x + 1 \right)}}{x^{2}}\right) = 0$$
More at x→-oo
The graph