Limits and Continuity

Substituting directly gives the indeterminate form \(\infty - \infty\), so multiply and divide by the conjugate, then divide the numerator and denominator by \(t > 0\):

\[\sqrt{t^{2} + t} - t = \frac{(t^{2} + t) - t^{2}}{\sqrt{t^{2} + t} + t} = \frac{t}{\sqrt{t^{2} + t} + t} = \frac{1}{\sqrt{1 + \frac{1}{t}} + 1}.\]

As \(t \to +\infty\), \(\frac{1}{t} \to 0\), so the limit is \(\frac{1}{1 + 1} = 0.5\). As a check, the expression is about 0.4988 at \(t = 100\).

Answer: 0.5

The logarithm of a product splits into a sum:

\[ \frac{\ln((x^{2}+1)\cos x)}{x^{2}} = \frac{\ln(1 + x^{2})}{x^{2}} + \frac{\ln(\cos x)}{x^{2}} \]

• Since \(\ln(1 + t) \approx t\) for small \(t\), the first term tends to 1.
• Since \(\cos x \approx 1 - \frac{x^{2}}{2}\), we get \(\ln(\cos x) \approx -\frac{x^{2}}{2}\), so the second term tends to \(-\frac{1}{2}\).

The limit is \(1 - \frac{1}{2} = 0.5\). L'Hôpital's rule agrees: differentiating the top and the bottom gives \(\frac{2x/(1 + x^{2}) - \tan x}{2x}\), which also tends to \(1 - \frac{1}{2}\).

Answer: 0.5