3
votes
Show that $f$ is strictly increasing on $\mathbb{R}_+.$
The following proof works for all $n \ge 2$.
$f$ is strictly convex on $\Bbb R$ because at most one of the $a_i$ can be equal to one.
$f$ has an absolute minimum at $x=0$ because
$$
f(x) \ge n \...
- 133k
3
votes
Find $ \lim_{x\to+\infty} \left( \frac{x^{2}+3}{3x^{2}+1} \right)^{x^{2}}=0 $
$$\displaystyle \lim_{x\to \infty} \left(\dfrac{x^2+3}{3x^2+1}\right)^{x^2}=\lim_{x\to \infty} e^{x^2\ln{\left(\dfrac{x^2+3}{3x^2+1}\right)}}\;.$$
Considering the exponent of the right-hand side ...
- 133
3
votes
Accepted
Find $ \lim_{x\to+\infty} \left( \frac{x^{2}+3}{3x^{2}+1} \right)^{x^{2}}=0 $
Firstly, analyse the behavior of the limit (correctly):
$$
\lim_{x\to+\infty}\frac{x^2+3}{3x^2+1}=\lim_{x\to+\infty}\frac{1+\dfrac{3}{x^2}}{3+\dfrac{1}{x^2}}=\frac{1+0}{3+0}=\frac{1}{3}\\
\lim_{x\to+\...
- 1,770
2
votes
Find $ \lim_{x\to+\infty} \left( \frac{x^{2}+3}{3x^{2}+1} \right)^{x^{2}}=0 $
For $x\ge 3 $ we have $$0<{x^2+3\over 3x^2+1}<{x^2+3\over 3x^2}={1\over 3}+{1\over x^2}< {1\over 2}$$ Thus
$$ 0<\left ({x^2+3\over 3x^2+1}\right )^{x^2}<{1\over 2^{x^2}},\quad x\ge 3$$ ...
- 46.9k
1
vote
One exponent for all numbers
For real numbers a,b,c such that $a>1$, if we assume that $a^b=a^c$, then we can take the logarithm of base $a$ on both sides to obtain $\log_a(a^b)=\log_a(a^c)$, then apply the logarithm to obtain ...
- 36
1
vote
One exponent for all numbers
Since $1^k=1$ for all possible $k$ then knowing $1^k = 1^m = 1^z = 1^2 = 1^w = 1^3=1^{-2578947839}=.... = 1$ can't possibly tell us anything about $k, m,z,w,....$
But if $a\ne 0, 1; a> 0$ then for ...
- 134k
1
vote
Find $ \lim_{x\to+\infty} \left( \frac{x^{2}+3}{3x^{2}+1} \right)^{x^{2}}=0 $
Should to proof
$$
\lim_{x\to+\infty}\left(\dfrac{x^2+3}{3x^2+1}\right)^{x^2}=0.
$$
Obviously we have
$\forall k\in\left(\frac13,1\right),
\exists N>0, \forall x > N,$ $$
0<\left(\dfrac{x^2+...
- 11
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