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.

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

Sure! I can help you with these problems. Let's solve them step by step:

1. To find the limit as x approaches infinity of (e^(-2x) + sin x), we need to analyze the behavior of each term separately.

- The term e^(-2x) represents an exponentially decaying function. As x approaches infinity, e^(-2x) approaches 0. This is because any positive number raised to a negative power approaches 0 as the exponent becomes larger.

- The term sin x represents a periodic function that oscillates between -1 and 1. As x approaches infinity, sin x does not approach any specific value since it continues to oscillate.

Therefore, the overall limit can be evaluated as the sum of the limits of each term:

lim x->infinity (e^(-2x) + sin x) = 0 + (oscillating term)

Since the oscillating term has no specific limit, the overall limit is indeterminate.

2. To find the derivative of sqrt(9 - x) using the limit process, we can apply the definition of the derivative:

f'(x) = lim h->0 (f(x + h) - f(x))/h

Let's substitute the function f(x) = sqrt(9 - x) into the derivative formula:

f'(x) = lim h->0 (sqrt(9 - (x + h)) - sqrt(9 - x))/h

To simplify this expression, we need to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the numerator:

f'(x) = lim h->0 ((sqrt(9 - (x + h)) - sqrt(9 - x))/h) * ((sqrt(9 - (x + h)) + sqrt(9 - x))/(sqrt(9 - (x + h)) + sqrt(9 - x)))

After simplifying, we can find the limit as h approaches 0:

f'(x) = lim h->0 (9 - (x + h) - 9)/h(sqrt(9 - (x + h)) + sqrt(9 - x))

Simplifying further, we get:

f'(x) = lim h->0 (-h)/(h(sqrt(9 - (x + h)) + sqrt(9 - x)))

Now we can cancel out the h term:

f'(x) = lim h->0 -1/(sqrt(9 - (x + h)) + sqrt(9 - x))

Finally, we substitute x back into the expression:

f'(x) = -1/(sqrt(9 - x) + sqrt(9 - x))

Simplifying the denominator:

f'(x) = -1/(2sqrt(9 - x))

So, the derivative of sqrt(9 - x) is -1/(2sqrt(9 - x)).

3. To find the limit as x approaches -infinity of (x + sqrt(x^2 + 2x)), we can analyze the behavior of each term separately.

- The term x represents a linear function. As x approaches -infinity, x also approaches -infinity.

- The term sqrt(x^2 + 2x) represents a function that involves a square root. Since we are taking the square root of a positive quantity (x^2 + 2x), the result will always be non-negative. Therefore, as x approaches -infinity, sqrt(x^2 + 2x) approaches 0.

Therefore, the overall limit can be evaluated as the sum of the limits of each term:

lim x-> -infinity (x + sqrt(x^2 + 2x)) = -infinity + 0 = -infinity

Hence, the limit as x approaches -infinity is -infinity.

4. To show that the equation e^x = 2 + 2x has a solution that is a negative number, we can use the intermediate value theorem.

First, let's rearrange the equation:

e^x - 2x - 2 = 0

Now, let's evaluate the function f(x) = e^x - 2x - 2 at two different points:

- When x = 0: f(0) = e^0 - 2(0) - 2 = 1 - 0 - 2 = -1
- When x = -1: f(-1) = e^(-1) - 2(-1) - 2 ≈ 0.27 + 2 + 2 ≈ 4.27

Since f(0) < 0 and f(-1) > 0, and the function f(x) is continuous, there must exist a value of x between 0 and -1 where f(x) = 0. In other words, there is at least one solution to the equation e^x = 2 + 2x that is a negative number.

However, to find the exact value of this solution, you may need to use numerical methods or approximations.