Math- Calculus

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(9-x) 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.

asked by Amber
  1. 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( √(9-x) - √(9-x-h) )/h , as h --->0
    = lim( √(9-x) - √(9-x-h) )/h * ( √(9-x) + √(9-x-h) )/( √(9-x) + √(9-x-h) )
    = lim ( 9-x - (9-x-h)/(h(√(9-x) + √(9-x-h) )
    = lim h/(h(√(9-x) + √(9-x-h) )
    = lim 1/(√(9-x) + √(9-x-h) ) , as h ---> 0
    = 1/(√(9-x) + √(9-x))
    = 1/(2√(9-x) )

    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

    posted by Reiny

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