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(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.
Math- Calculus -
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..
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.
As you can see, there are actually two solutions, one x is positive, the other is negative.
Let's actually solve the equation
notice that x = -.768039 and x = +1.67835
LS = e^1.67835 = 5.35671
RS = 2(1.67835) + 2 = 5.3567 , close enough