Math Calculus
posted by Amber .
I need help with these problems, I cannot find a similar example to help me in the book:
1. Find lim x>infinity (e^(2x) + sin x).
2. Find the derivative of sqrt(9x) using the limit process.
3. Find lim x> infinity (x + sqrt(x^2 + 2x)).
4. Show that the equation e^x = 2+2x has a solution that is a negative number.

Math Calculus 
Reiny
1. done by just getting a "feel" for the numbers
e^(2x) = 1/e^(2x)
so as x > infinitity, the denominator gets huge and the result approaches zero
as for the sinx it simply runs up and down between 1 and 1 no matter what x is or how large x is
So the lim(e^(2x) + sinx) does not reach an actual value, but simply fluctuages between 1 and 1
e.g. x = 500000
value = e^(1000000) + sin(50000)
= "almost" zero + .1778..
for x = 500002
I get "zero" + sin500002 = .9688..
2.
dy/dx = lim( √(9x)  √(9xh) )/h , as h >0
= lim( √(9x)  √(9xh) )/h * ( √(9x) + √(9xh) )/( √(9x) + √(9xh) )
= lim ( 9x  (9xh)/(h(√(9x) + √(9xh) )
= lim h/(h(√(9x) + √(9xh) )
= lim 1/(√(9x) + √(9xh) ) , as h > 0
= 1/(√(9x) + √(9x))
= 1/(2√(9x) )
the x>0 should be included in each line except the last two, I was just being lazy
3. since there is no denominator, nor are we taking √ of a negative, there should be no problem here,
Again, just get a feel for the numbers
look at the √(x^2 + 2x)
as gets larger into the negatives, x^2 makes it positive
and x^2 gets bigger much faster than 2x
so the x^2 eventually leaves the 2x behind and taking √ brings us back to x
so we have lim (x + x) or lim 0, which is 0
4. e^x = 2+2x
let y1= e^x and let y2 = 2x+2, the latter is a straight line
let's graph both of these.
http://www.wolframalpha.com/input/?i=plot+y+%3D+e%5Ex+%2C+y%3D2x%2B2
As you can see, there are actually two solutions, one x is positive, the other is negative.
Let's actually solve the equation
http://www.wolframalpha.com/input/?i=solve+e%5Ex%3D2x%2B2
notice that x = .768039 and x = +1.67835
testing:
LS = e^1.67835 = 5.35671
RS = 2(1.67835) + 2 = 5.3567 , close enough
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