\( \lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+x}-\sqrt{x^{2}-x}\right) \quad \) (c) \( \quad \lim _{x \rightarrow \infty}\left[\frac{\sqrt{y+1}+\sqrt{y-1}}{y}\right] \)
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To solve the limit \( \lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+x}-\sqrt{x^{2}-x}\right) \), we can first factor out \( \sqrt{x^2} \) from each square root. We rewrite it as follows: \[ \sqrt{x^{2}+x}=\sqrt{x^{2}(1+\frac{1}{x})} = x\sqrt{1+\frac{1}{x}} \] \[ \sqrt{x^{2}-x}=\sqrt{x^{2}(1-\frac{1}{x})} = x\sqrt{1-\frac{1}{x}} \] Now our limit becomes: \[ \lim_{x \to \infty} \left( x\sqrt{1+\frac{1}{x}} - x\sqrt{1-\frac{1}{x}} \right) = \lim_{x \to \infty} x\left(\sqrt{1+\frac{1}{x}} - \sqrt{1-\frac{1}{x}}\right) \] As \( x \to \infty \), both \( \sqrt{1+\frac{1}{x}} \) and \( \sqrt{1-\frac{1}{x}} \) approach 1. However, we can use the binomial expansion for small values of the arguments: \[ \sqrt{1+\frac{1}{x}} \approx 1 + \frac{1}{2x}, \quad \sqrt{1-\frac{1}{x}} \approx 1 - \frac{1}{2x} \] So, we have: \[ \sqrt{1+\frac{1}{x}} - \sqrt{1-\frac{1}{x}} \approx \left(1 + \frac{1}{2x}\right) - \left(1 - \frac{1}{2x}\right) = \frac{1}{x} \] Now substitute back: \[ \lim_{x \to \infty} x\left(\frac{1}{x}\right) = \lim_{x \to \infty} 1 = 1 \] Thus, the solution to the first limit is: \[ \lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+x}-\sqrt{x^{2}-x}\right) = 1 \] --- For the second limit \( \lim_{y \to \infty} \left[\frac{\sqrt{y+1} + \sqrt{y-1}}{y}\right] \), we will simplify the expression similarly by dividing the terms in the numerator by \( y \): \[ = \lim_{y \to \infty} \left[\frac{\sqrt{y(1 + \frac{1}{y})} + \sqrt{y(1 - \frac{1}{y})}}{y}\right] \] Rearranging gives: \[ = \lim_{y \to \infty} \left[\frac{\sqrt{y}(\sqrt{1 + \frac{1}{y}} + \sqrt{1 - \frac{1}{y}})}{y}\right] = \lim_{y \to \infty} \frac{\sqrt{1 + \frac{1}{y}} + \sqrt{1 - \frac{1}{y}}}{\sqrt{y}} \] As \( y \to \infty \), \( \sqrt{1 + \frac{1}{y}} \) and \( \sqrt{1 - \frac{1}{y}} \) approach 1. Therefore, we have: \[ \frac{1 + 1}{\sqrt{y}} = \frac{2}{\sqrt{y}} \to 0 \text{ as } y \to \infty \] Thus, the solution to the second limit is: \[ \lim_{y \to \infty} \left[\frac{\sqrt{y+1} + \sqrt{y-1}}{y}\right] = 0 \] In summary: 1. The first limit equals 1. 2. The second limit equals 0.
