# How do you find the limit of ((t^2)+(5t)) / (cosh(t)-1) as t approaches 0?

Jul 12, 2016

${\lim}_{t \to {0}^{-}} \frac{{t}^{2} + 5 t}{\cosh \left(t\right) - 1} = - \infty$

${\lim}_{t \to {0}^{+}} \frac{{t}^{2} + 5 t}{\cosh \left(t\right) - 1} = \infty$

#### Explanation:

${\lim}_{t \to 0} \frac{{t}^{2} + 5 t}{\cosh \left(t\right) - 1}$

currently this is $\frac{0}{0}$, indeterminate, so we can apply Lhopital's Rule

$= {\lim}_{t \to 0} \frac{2 t + 5}{\sinh \left(t\right)}$

= lim_{t to 0} 2 (t) / (sinh(t)) +5 / (sinh(t)

From L'Hopital: ${\lim}_{t \to 0} 2 \frac{t}{\sinh \left(t\right)} = 2 {\lim}_{t \to 0} \frac{1}{\cosh \left(t\right)} = 2$

implies 2 + lim_{t to 0} 5 / (sinh(t)

Now,  lim_{t to 0} 5 / (sinh(t) is 2 sided limit ie

$= {\lim}_{t \to {0}^{-}} \frac{5}{\sinh} \left(t\right) = - \infty$

$= {\lim}_{t \to {0}^{+}} \frac{5}{\sinh} \left(t\right) = \infty$

so

${\lim}_{t \to {0}^{-}} \frac{{t}^{2} + 5 t}{\cosh \left(t\right) - 1} = - \infty$

${\lim}_{t \to {0}^{+}} \frac{{t}^{2} + 5 t}{\cosh \left(t\right) - 1} = \infty$