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