How do you find #lim sqrt(2x+3)-sqrtx# as #x->oo#?

As a result,

To find the limit of sqrt(2x+3) - sqrt(x) as x approaches infinity, we can use algebraic manipulation.

First, we multiply the expression by its conjugate, which is sqrt(2x+3) + sqrt(x). This will help us eliminate the square roots in the numerator.

By multiplying the numerator and denominator by the conjugate, we get:

lim (sqrt(2x+3) - sqrt(x)) * (sqrt(2x+3) + sqrt(x)) / (sqrt(2x+3) + sqrt(x)) as x approaches infinity

Simplifying the numerator using the difference of squares, we have:

lim (2x+3) - x / (sqrt(2x+3) + sqrt(x)) as x approaches infinity

Simplifying further, we get:

lim x + 3 / (sqrt(2x+3) + sqrt(x)) as x approaches infinity

As x approaches infinity, the terms involving x dominate the expression. Therefore, we can ignore the constant term 3.

Thus, the limit becomes:

lim x / (sqrt(2x+3) + sqrt(x)) as x approaches infinity

To simplify this further, we can divide both the numerator and denominator by sqrt(x):

lim (x / sqrt(x)) / (sqrt(2x+3) / sqrt(x) + sqrt(x) / sqrt(x)) as x approaches infinity

Simplifying, we get:

lim sqrt(x) / (sqrt(2 + 3/x) + 1) as x approaches infinity

As x approaches infinity, 3/x approaches 0. Therefore, the expression becomes:

lim sqrt(x) / (sqrt(2 + 0) + 1) as x approaches infinity

Simplifying further, we have:

lim sqrt(x) / (sqrt(2) + 1) as x approaches infinity

Since the square root of x grows without bound as x approaches infinity, the limit becomes:

infinity / (sqrt(2) + 1) as x approaches infinity

Therefore, the limit of sqrt(2x+3) - sqrt(x) as x approaches infinity is infinity divided by (sqrt(2) + 1).