Math  /  Calculus

QuestionFind the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. limt0e5t1sin(t)\lim _{t \rightarrow 0} \frac{e^{5 t}-1}{\sin (t)} Need Help? Read It Watch It

Studdy Solution
Evaluate the limit by substituting t=0t = 0:
limt05e5tcos(t)=5e0cos(0)=511=5 \lim_{t \rightarrow 0} \frac{5e^{5t}}{\cos(t)} = \frac{5e^{0}}{\cos(0)} = \frac{5 \cdot 1}{1} = 5
The value of the limit is:
5 \boxed{5}

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