Answer:
![$\lim_{x \to 0} \frac{2\cos(x)+1}{\sin^2 (x) } = \infty$](https://tex.z-dn.net/?f=%24%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B2%5Ccos%28x%29%2B1%7D%7B%5Csin%5E2%20%28x%29%20%7D%20%3D%20%5Cinfty%24)
and the limit does not exist
Step-by-step explanation:
We are given the following limit
![$ \lim_{x \to 0} \frac{\sin(2x)+\sin(x)}{\sin^3 (x)} $](https://tex.z-dn.net/?f=%24%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csin%282x%29%2B%5Csin%28x%29%7D%7B%5Csin%5E3%20%28x%29%7D%20%24)
We can't just merely consider
![$ \lim_{x \to 0} f(x)=g(x) $](https://tex.z-dn.net/?f=%24%20%5Clim_%7Bx%20%5Cto%200%7D%20f%28x%29%3Dg%28x%29%20%24)
and then solve
![$ \lim_{x \to 0} \frac{\sin(2x)+\sin(x)}{\sin^3 (x)} = \frac{\sin(0)+\sin(0)}{\sin^3 (0)} =\frac{0}{0} =0$](https://tex.z-dn.net/?f=%24%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csin%282x%29%2B%5Csin%28x%29%7D%7B%5Csin%5E3%20%28x%29%7D%20%3D%20%5Cfrac%7B%5Csin%280%29%2B%5Csin%280%29%7D%7B%5Csin%5E3%20%280%29%7D%20%3D%5Cfrac%7B0%7D%7B0%7D%20%3D0%24)
This is an indeterminate form. Wrong.
=========================================
We have the identity
, therefore we can substitute it in the limit
![$ \lim_{x \to 0} \frac{\sin(2x)+\sin(x)}{\sin^3 (x)} = \lim_{x \to 0} \frac{2\sin(x)\cos(x)+\sin(x)}{\sin^3 (x)} $](https://tex.z-dn.net/?f=%24%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csin%282x%29%2B%5Csin%28x%29%7D%7B%5Csin%5E3%20%28x%29%7D%20%3D%20%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B2%5Csin%28x%29%5Ccos%28x%29%2B%5Csin%28x%29%7D%7B%5Csin%5E3%20%28x%29%7D%20%24)
then
![$ \lim_{x \to 0} \frac{2\sin(x)\cos(x)+\sin(x)}{\sin^3 (x)} = \lim_{x \to 0} \frac{\sin(x)(2\cos(x)+1)}{\sin^3 (x)} = \boxed{ \lim_{x \to 0} \frac{2\cos(x)+1}{\sin^2 (x)} }$](https://tex.z-dn.net/?f=%24%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B2%5Csin%28x%29%5Ccos%28x%29%2B%5Csin%28x%29%7D%7B%5Csin%5E3%20%28x%29%7D%20%3D%20%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B%5Csin%28x%29%282%5Ccos%28x%29%2B1%29%7D%7B%5Csin%5E3%20%28x%29%7D%20%3D%20%5Cboxed%7B%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B2%5Ccos%28x%29%2B1%7D%7B%5Csin%5E2%20%28x%29%7D%20%7D%24)
Now, differently from the first case,
we have
as
because ![\cos(0)=1](https://tex.z-dn.net/?f=%5Ccos%280%29%3D1)
On the other hand,
as
because ![\sin(0)=0](https://tex.z-dn.net/?f=%5Csin%280%29%3D0)
In fact,
![$\lim_{x \to 0} \frac{2\cos(x)+1}{\sin^2 (x) } = \lim_{x \to 0} \frac{1}{\sin^2 (x) }\cdot (2\cos(x)+1) = \infty \cdot 3 = \infty $](https://tex.z-dn.net/?f=%24%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B2%5Ccos%28x%29%2B1%7D%7B%5Csin%5E2%20%28x%29%20%7D%20%3D%20%5Clim_%7Bx%20%5Cto%200%7D%20%5Cfrac%7B1%7D%7B%5Csin%5E2%20%28x%29%20%7D%5Ccdot%20%282%5Ccos%28x%29%2B1%29%20%3D%20%5Cinfty%20%5Ccdot%20%203%20%3D%20%5Cinfty%20%24)
Answer:
The points of discontinuity are: x=5 and x=1. The point of discontinuity x=5 can be removable.
Step-by-step explanation:
The points of discontinuity are those points where the function is not defined. To find such points, we should factorize the denominator of the function needs to be factorized.
![y= (x-5)/(x^2 - 6x + 5)](https://tex.z-dn.net/?f=y%3D%20%28x-5%29%2F%28x%5E2%20-%206x%20%2B%205%29)
Considering the quadratic equation of the form ax^2+bx+c =0, then using the quadratic (see attached image), where a=, b=- and c=5, we have that the roots are:
x=5 and x=1.
If we simplify the fraction, by removing the term (x-5) from both the numerator and the denominator, we get:
, so we removed the point of discontinuity x=5.
Answer:
b should increase
Step-by-step explanation:
The answer is 15 , because First you add them all together (75) then divide that number by the number of numbrs (15) which will give you the mean
All you need to do is 16 so she needs to work 16 hours to get $115.