Odd function: f(-x) = -f(x)
Let two odd functions be f(x) and g(x)
Let k(x) = f(x) + g(x)
Then k(-x) = f(-x) + g(-x) = -(f(x) + g(x)) = -k(x)
So k(x) is odd too.
Answer:
Yep, they all match:
1. Correct
2. Correct
3. Correct
4. Correct
Step-by-step explanation:
1.
√2 • √3 • √3 = 4.242640687119285
3√2 = 4.242640687119285
2.
√2 • √3 • √-3 = 4.24264069i
3i√2 = 4.24264069i
3.
√-2 • √-3 • √-3 = -4.24264069i
-3√2 = -4.24264069i
4.
√2 • √-3 • √-3 = -4.24264069i
-3i√2 = -4.24264069i
The answer is 1.5. Or 1 1/2 in fraction form. You need to divide for this problem
0 divided by 2 is 0.
You can use the multiplicative identity property: the product of any number and zero is still zero.
Hope this helped :)
Answer:
We want to find:
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D)
Here we can use Stirling's approximation, which says that for large values of n, we get:

Because here we are taking the limit when n tends to infinity, we can use this approximation.
Then we get.
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7B%5Csqrt%7B2%2A%5Cpi%2An%7D%20%2A%28%5Cfrac%7Bn%7D%7Be%7D%20%29%5En%7D%20%7D%7Bn%7D%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7Bn%7D%7Be%2An%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D)
Now we can just simplify this, so we get:
![\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Be%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D%20%5C%5C)
And we can rewrite it as:

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus:
