Questions: Find the limit of the following. lim as x approaches infinity of (sqrt(x^2+25)-sqrt(x^2-16)) lim as x approaches infinity of (sqrt(x^2+25)-sqrt(x^2-16)) = (Simplify your answer.)

Solution Steps

To find the limit of the expression as \( x \) approaches infinity, we can use algebraic manipulation. Specifically, we can multiply and divide by the conjugate of the expression to simplify it. This will help eliminate the square roots and make it easier to evaluate the limit.

We are tasked with finding the limit of the expression as \( x \) approaches infinity: \[ \lim_{x \to \infty} \left( \sqrt{x^2 + 25} - \sqrt{x^2 - 16} \right) \]

To simplify the expression, multiply and divide by the conjugate: \[ \frac{\left( \sqrt{x^2 + 25} - \sqrt{x^2 - 16} \right) \left( \sqrt{x^2 + 25} + \sqrt{x^2 - 16} \right)}{\sqrt{x^2 + 25} + \sqrt{x^2 - 16}} \]

The numerator becomes: \[ (\sqrt{x^2 + 25})^2 - (\sqrt{x^2 - 16})^2 = (x^2 + 25) - (x^2 - 16) = 41 \] Thus, the expression simplifies to: \[ \frac{41}{\sqrt{x^2 + 25} + \sqrt{x^2 - 16}} \]

As \( x \to \infty \), the dominant term in the denominator is \( 2x \), so: \[ \lim_{x \to \infty} \frac{41}{\sqrt{x^2 + 25} + \sqrt{x^2 - 16}} = \lim_{x \to \infty} \frac{41}{2x} = 0 \]

Final Answer