How do you determine the limit of #sinh(2x)/e^(3x)# as x approaches infinity?

Answer 1

#= 0#

#lim_(x to oo) sinh(2x)/e^(3x)#
#= lim_(x to oo) (e^(2x) - e^(-2x))/(2e^(3x))#
#= lim_(x to oo) (e^(-x) - e^(-5x))/(2e^(0))#
#= lim_(x to oo) (e^(-x) - e^(-5x))/(2)#
#= (lim_(x to oo)e^(-x) - e^(-5x))/(2)#
#= 0#
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Answer 2

To determine the limit of sinh(2x)/e^(3x) as x approaches infinity, we can use the concept of limits. By applying L'Hôpital's Rule, we can differentiate the numerator and denominator separately with respect to x. Taking the derivative of sinh(2x) gives us 2cosh(2x), and the derivative of e^(3x) is 3e^(3x).

Now, we can rewrite the limit as the limit of (2cosh(2x))/(3e^(3x)) as x approaches infinity. As x approaches infinity, both cosh(2x) and e^(3x) also approach infinity. Therefore, we have an indeterminate form of infinity/infinity.

Applying L'Hôpital's Rule again, we differentiate the numerator and denominator once more. The derivative of 2cosh(2x) is 4sinh(2x), and the derivative of 3e^(3x) is 9e^(3x).

Now, we have the limit of (4sinh(2x))/(9e^(3x)) as x approaches infinity. As x approaches infinity, sinh(2x) and e^(3x) both still approach infinity.

Applying L'Hôpital's Rule for the third time, we differentiate the numerator and denominator again. The derivative of 4sinh(2x) is 8cosh(2x), and the derivative of 9e^(3x) is 27e^(3x).

Now, we have the limit of (8cosh(2x))/(27e^(3x)) as x approaches infinity. As x approaches infinity, cosh(2x) and e^(3x) both still approach infinity.

Since we have applied L'Hôpital's Rule three times and still have an indeterminate form of infinity/infinity, we can continue this process indefinitely.

In conclusion, the limit of sinh(2x)/e^(3x) as x approaches infinity does not exist.

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Answer from HIX Tutor

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.

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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