lim(sqrt(x+1))/(sqrt(4x-1)) as x approaches infinity.

intuitive approach:

as x ---> ∞ √(x+1) --> √x
and √(4x+1) ---> √(4x)

so
lim(√(x+1))/(√(4x-1)) as x ---> ∞
= lim √x/√(4x)
= √(1/4) = 1/2 , as ---> ∞

Thank you! I think I just overwhelmed with the idea of infinity. :)

To find the limit as x approaches infinity for the expression lim(sqrt(x+1))/(sqrt(4x-1)), we can use a technique called rationalizing the denominator.

First, let's rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is sqrt(4x-1) + sqrt(x+1).

Multiply the numerator and denominator by the conjugate:
lim(sqrt(x+1))/(sqrt(4x-1)) * (sqrt(4x-1) + sqrt(x+1))/(sqrt(4x-1) + sqrt(x+1))

Now, let's simplify the expression:
lim((sqrt(x+1))*(sqrt(4x-1) + sqrt(x+1)))/((sqrt(4x-1))*(sqrt(4x-1) + sqrt(x+1)))

Distribute and simplify:
lim(sqrt(x+1)*sqrt(4x-1) + sqrt(x+1)*sqrt(x+1))/(sqrt(4x-1)*sqrt(4x-1) + sqrt(4x-1)*sqrt(x+1))

Now, we can simplify the expression further:
lim(sqrt((x+1)*(4x-1)) + (x+1))/(sqrt((4x-1)*(4x-1)) + sqrt(4x-1)*sqrt(x+1))

Now, let's simplify inside the square roots:
lim(sqrt(4x^2 + 3x)) + (x+1))/(sqrt(16x^2 - 8x + 1) + sqrt(4x^2 - x))

As x approaches infinity, the highest exponent (x^2) dominates the expression. So, let's ignore lower order terms and simplify further:
lim(sqrt(4x^2)) / (sqrt(16x^2)) = 2x / (4x) = 1/2

Therefore, the limit of sqrt(x+1))/(sqrt(4x-1) as x approaches infinity is 1/2.