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:
[tex] \lim_{x \to 2}( \frac{f'( \sqrt{x+7}+ \sqrt[3]{4x+19} ) }{f'(x-2)})[/tex]
Which is equal to:
[tex] \lim_{x \to 2}( \frac{f'((x+7)^{1/2} -(4x+19)^{1/3} ) }{f'(x-2)})[/tex]
Which is then equal to:
[tex] \lim_{x \to 2}( \frac{((1/2)(x+7)^{-1/2} -(1/3)(4)(4x+19)^{-2/3} ) }{1})[/tex]
If you then substitute back in 2, you get:
[tex]( \frac{((1/2)(1/3)-(4/3)(1/9) }{1}) = (1/6)-(4/27)=(9/54)-(8/54) = 1/54[/tex]
Then in the next question you have to separate and integrate.
The original equation is:
[tex] \frac{dy}{dx} = \frac{3}{ \sqrt{5x-2} } [/tex]
Then you separate and integrate so:
[tex] \int\limits^._. {1} \, dy = \int\limits^ . _ . {3(5x-2)^{-1/2} \, dx [/tex]
Then you would use u substitution on the right side with 5x-2 as u, and 5 as du.
Next you would have:
[tex]y = 3/5( \int\limits^._. {u}^{-1/2} \, du) = \frac{6u^{1/2}}{5}+C [/tex]
[tex]y=\frac{6(5x-2)^{1/2}}{5}+C[/tex]
To find C, you plug in x = 3, and y = 6
So:
[tex]6=\frac{6(13)^{1/2}}{5}+C[/tex]
C = 1.6733
And the equation would be:
[tex]y = \frac{6(5x-2)^{1/2}}{5}+1.6733[/tex]
