Let

\[ B(t) = 4 \log _{2}(5t - 3) - 1 \]
\[ \lim \limits _{t \to \infty } B(t) = \answer {\infty } \]
\[ \lim \limits _{t \to \frac {3}{5}^+} B(t) = \answer {-1} \]

Let

\[ h(z) = 4 \ln (-5z - 3) - 1 \]
\[ \lim \limits _{z \to -\infty } h(z) = \answer {\infty } \]
\[ \lim \limits _{z \to \frac {-3}{5}^-} h(z) = \answer {-\infty } \]

Let

\[ f(y) = -4 \log _{5}(5y - 3) - 1 \]
\[ \lim \limits _{y \to \infty } f(y) = \answer {-\infty } \]
\[ \lim \limits _{y \to \frac {-3}{5}^+} f(y) = \answer {\infty } \]

Let

\[ H(x) = -4 \ln (-5x - 3) - 1 \]
\[ \lim \limits _{x \to \infty } H(x) = \answer {\infty } \]
\[ \lim \limits _{x \to \frac {-3}{5}^-} H(x) = \answer {-\infty } \]
2025-01-07 02:18:31