De l'Hospital rule applies to undetermined forms like
![\dfrac{0}{0},\quad\dfrac{\infty}{\infty}](https://tex.z-dn.net/?f=%5Cdfrac%7B0%7D%7B0%7D%2C%5Cquad%5Cdfrac%7B%5Cinfty%7D%7B%5Cinfty%7D)
If we evaluate your limit directly, we have
![\displaystyle \lim_{x\to 0}(1+2x)^{-\frac{3}{x}} = 1^\infty](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%7D%281%2B2x%29%5E%7B-%5Cfrac%7B3%7D%7Bx%7D%7D%20%3D%201%5E%5Cinfty)
which is neither of the two forms covered by the theorem.
So, in order to apply it, we need to write the limit as follows: we start with
![f(x)=(1+2x)^{-\frac{3}{x}}](https://tex.z-dn.net/?f=f%28x%29%3D%281%2B2x%29%5E%7B-%5Cfrac%7B3%7D%7Bx%7D%7D)
Using the identity
, we can rewrite the function as
![f(x)=e^{\log\left((1+2x)^{-\frac{3}{x}}\right)}](https://tex.z-dn.net/?f=f%28x%29%3De%5E%7B%5Clog%5Cleft%28%281%2B2x%29%5E%7B-%5Cfrac%7B3%7D%7Bx%7D%7D%5Cright%29%7D)
Using the rule
, we have
![f(x)=e^{-\frac{3}{x}\log(1+2x)}](https://tex.z-dn.net/?f=f%28x%29%3De%5E%7B-%5Cfrac%7B3%7D%7Bx%7D%5Clog%281%2B2x%29%7D)
Since the exponential function
is continuous, we have
![\displaystyle \lim_{x\to 0} e^{f(x)} = e^{\lim_{x\to 0} f(x)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%7D%20e%5E%7Bf%28x%29%7D%20%3D%20e%5E%7B%5Clim_%7Bx%5Cto%200%7D%20f%28x%29%7D)
In other words, we can focus on the exponent alone to solve the limit. So, we're focusing on
![\displaystyle \lim_{x\to 0} -\frac{3}{x}\log(1+2x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%7D%20-%5Cfrac%7B3%7D%7Bx%7D%5Clog%281%2B2x%29%20)
Which we can rewrite as
![\displaystyle \lim_{x\to 0} -\frac{3}{x}\log(1+2x) = -3\lim_{x\to 0}\frac{\log(1+2x)}{x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%7D%20-%5Cfrac%7B3%7D%7Bx%7D%5Clog%281%2B2x%29%20%3D%20-3%5Clim_%7Bx%5Cto%200%7D%5Cfrac%7B%5Clog%281%2B2x%29%7D%7Bx%7D)
Now the limit comes in the form 0/0, so we can apply the theorem: we derive both numerator and denominator to get
![\displaystyle -3\lim_{x\to 0}\frac{\log(1+2x)}{x} = -3 \lim_{x\to 0}\dfrac{\frac{2}{1+2x}}{1} = -3\cdot 2 = -6](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-3%5Clim_%7Bx%5Cto%200%7D%5Cfrac%7B%5Clog%281%2B2x%29%7D%7Bx%7D%20%3D%20-3%20%5Clim_%7Bx%5Cto%200%7D%5Cdfrac%7B%5Cfrac%7B2%7D%7B1%2B2x%7D%7D%7B1%7D%20%3D%20-3%5Ccdot%202%20%3D%20-6)
So, the limit of the exponent is -6, which implies that the whole expression tends to
![e^{-6}=\dfrac{1}{e^6}](https://tex.z-dn.net/?f=e%5E%7B-6%7D%3D%5Cdfrac%7B1%7D%7Be%5E6%7D)
Hello.
In order to see if the ratios are equal, we have to simplify both of them.
The first ratio is 8:4.
Divide 8 and 4 by 4:
2:1
Now, simplify the second ratio:
16:64
4:16
1:4
Are 2:1 and 1:4 equal? No.
Therefore, 8:4 and 16:64 are not equal ratios.
I hope this helps.
Have a nice day.
![\boxed{imperturbability}](https://tex.z-dn.net/?f=%5Cboxed%7Bimperturbability%7D)
Answer:
I think there would be 86 blue marbles
x=4(106 - x ) + 6
x= 424 - 4x + 6
5x= 430
x= 86 blue marbles
Tell me if I'm wrong, hope this helps.
The best and most correct answer among the choices provided by your question is the third choice or letter C.
A diagonal is the term that describes the dotted line in the quadrilateral.
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