Final answer:
The limit of the function as t approaches 0 is found using L'Hopital's Rule and simplifies to 7.
Step-by-step explanation:
The limit of the function as t approaches 0 of e^(7t) - 1 over sin(t) can be found using L'Hopital's Rule, which states that if the limit of f(t)/g(t) as t approaches a value gives an indeterminate form such as 0/0, then the limit of f(t)/g(t) equals the limit of f'(t)/g'(t) as t approaches that value, provided the latter limit exists.
To apply L'Hopital's Rule, we first differentiate both the numerator and the denominator with respect to t. The derivative of e^(7t) with respect to t is 7e^(7t), and the derivative of sin(t) with respect to t is cos(t). Thus, applying L'Hopital's Rule:
lim t→0 (7e^(7t)) / cos(t) = 7e^(7*0) / cos(0) = 7 / 1 = 7.
Therefore, the limit of the original function as t approaches 0 is 7.