2.7. \( \lim _{t \rightarrow \infty} \frac{\sqrt{t^{2}+16}}{t+1} \)

To solve the limit \( \lim _{t \rightarrow \infty} \frac{\sqrt{t^{2}+16}}{t+1} \), let's simplify the expression. First, we divide the numerator and denominator by \( t \), the highest power of \( t \) in the denominator: \[ \lim_{t \rightarrow \infty} \frac{\sqrt{t^{2}+16}}{t+1} = \lim_{t \rightarrow \infty} \frac{\sqrt{t^{2}\left(1+\frac{16}{t^{2}}\right)}}{t\left(1+\frac{1}{t}\right)} \] This simplifies to: \[ \lim_{t \rightarrow \infty} \frac{t\sqrt{1+\frac{16}{t^{2}}}}{t\left(1+\frac{1}{t}\right)} = \lim_{t \rightarrow \infty} \frac{\sqrt{1+\frac{16}{t^{2}}}}{1+\frac{1}{t}} \] As \( t \rightarrow \infty \), \( \frac{16}{t^{2}} \rightarrow 0 \) and \( \frac{1}{t} \rightarrow 0 \). Thus, the limit reduces to: \[ \frac{\sqrt{1+0}}{1+0} = \frac{1}{1} = 1 \] Therefore, the final answer is: \[ \lim _{t \rightarrow \infty} \frac{\sqrt{t^{2}+16}}{t+1} = 1 \]