Question

Comprehensive resolution
Please!!!!!

Answer 1

First lets look at one of the limit problems. I will do number 48, but the same principles apply to all of them. Because when you first input 2, you get the limit of 0/0 and is indeterminate, you would go ahead and use l'Hopitals rule which is defined as follows:

`\lim_(x \to 2)( \frac{f'( √(x+7)+ \sqrt[3]{4x+19} ) }{f'(x-2)})`
Which is equal to:

`\lim_(x \to 2)( (f'((x+7)^(1/2) -(4x+19)^(1/3) ) )/(f'(x-2)))`
Which is then equal to:

`\lim_(x \to 2)( (((1/2)(x+7)^(-1/2) -(1/3)(4)(4x+19)^(-2/3) ) )/(1))`
If you then substitute back in 2, you get:

`( (((1/2)(1/3)-(4/3)(1/9) )/(1)) = (1/6)-(4/27)=(9/54)-(8/54) = 1/54`

Then in the next question you have to separate and integrate.
The original equation is:


`(dy)/(dx) = (3)/( √(5x-2) )`

Then you separate and integrate so:



`\int\limits^._. {1} \, dy = \int\limits^ . _ . {3(5x-2)^(-1/2) \, dx`

Then you would use u substitution on the right side with 5x-2 as u, and 5 as du.
Next you would have:


`y = 3/5( \int\limits^._. {u}^(-1/2) \, du) = (6u^(1/2))/(5)+C`


`y=(6(5x-2)^(1/2))/(5)+C`

To find C, you plug in x = 3, and y = 6

So:

`6=(6(13)^(1/2))/(5)+C`

C = 1.6733

And the equation would be:


`y = (6(5x-2)^(1/2))/(5)+1.6733`