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

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To find the limit of ((t^2)+(5t)) / (cosh(t)-1) as t approaches 0, we can use L'Hôpital's rule. Taking the derivative of the numerator and denominator separately, we get (2t + 5) / sinh(t). Evaluating this expression as t approaches 0, we find that the limit is 5/1, which simplifies to 5.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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