Evaluate the limit. as x approaches infinity, lim sqrt(x^2+6x+3)-x
sqrt(x^2+6x+3) =
x sqrt(1 + 6/x + 3/x^2)
Using the series expansion:
(1+y)^p = 1 + p y + p (p-1)/2 y^2 + ...
for p = 1/2 gives:
sqrt(1 + 6/x + 3/x^2) = 1 + 3/x +
O(1/x^2)
Where the O(1/x^2) means that there exists an R and a constant c such that for x larger than R, the difference between the square root and (1+3/x) becomes less than c/x^2.
We then have:
x sqrt(1 + 6/x + 3/x^2) - x =
3 + O(1/x)
This means that the limit for x to infinity is 3.
To evaluate the limit as x approaches infinity for the expression sqrt(x^2+6x+3) - x, we can use algebraic manipulations to simplify the expression.
Step 1: Rewrite the expression with a common denominator.
lim sqrt(x^2+6x+3) - x = lim (sqrt(x^2+6x+3) - x) * (sqrt(x^2+6x+3) + x) / (sqrt(x^2+6x+3) + x)
Step 2: Simplify the numerator.
lim (sqrt(x^2+6x+3))^2 - x^2 / (sqrt(x^2+6x+3) + x)
Step 3: Expand and cancel out the x^2 terms.
lim (x^2 + 6x + 3 - x^2) / (sqrt(x^2+6x+3) + x)
Step 4: Simplify the expression.
lim (6x + 3) / (sqrt(x^2+6x+3) + x)
Step 5: Divide both numerator and denominator by x.
lim (6 + 3/x) / (sqrt(1+6/x+3/x^2) + 1)
Step 6: Take the limit as x approaches infinity.
As x approaches infinity, 3/x approaches 0, and 6/x^2 approaches 0, and 1+6/x+3/x^2 approaches 1.
lim (6 + 0) / (sqrt(1+0+0) + 1)
Step 7: Simplify the expression further.
lim 6 / (sqrt(1) + 1) = 6 / (1 + 1)
Step 8: Evaluate the final result.
lim 6 / 2 = 3
Therefore, the limit as x approaches infinity for sqrt(x^2+6x+3) - x is 3.
To evaluate the limit as x approaches infinity for the given expression, lim(sqrt(x^2+6x+3)-x), we can simplify the expression first.
Step 1: Start by rewriting the expression.
lim(sqrt(x^2+6x+3)-x) = lim(sqrt(x^2+6x+3)-x) * (sqrt(x^2+6x+3)+x) / (sqrt(x^2+6x+3)+x)
Step 2: Simplify the expression by using the conjugate.
= lim((x^2+6x+3)-x^2) / (sqrt(x^2+6x+3)+x)
Step 3: Simplify further.
= lim(6x+3) / (sqrt(x^2+6x+3)+x)
Step 4: As x approaches infinity, the numerator 6x+3 goes to infinity, and the denominator sqrt(x^2+6x+3)+x also goes to infinity.
Therefore, the limit as x approaches infinity of sqrt(x^2+6x+3)-x is equal to infinity divided by infinity, which is an indeterminate form.
To further evaluate this limit, more information or a specific function/formula is needed.