# How do you find the limit of #(e^(3x) - e^(5x))/(x)# as x approaches 0?

To find the limit of ( \frac{e^{3x} - e^{5x}}{x} ) as ( x ) approaches 0, we can use L'Hôpital's Rule.

- First, we evaluate the function at ( x = 0 ) to see if it is an indeterminate form.
- If it is an indeterminate form, we take the derivative of the numerator and the derivative of the denominator separately and then evaluate the limit again.

Let's proceed with the steps:

- Evaluate the function at ( x = 0 ): [ \lim_{x \to 0} \frac{e^{3x} - e^{5x}}{x} = \frac{e^{3(0)} - e^{5(0)}}{0} = \frac{e^0 - e^0}{0} = \frac{1 - 1}{0} = \frac{0}{0} ]

Since we obtained the indeterminate form ( \frac{0}{0} ), we can apply L'Hôpital's Rule.

- Take the derivative of the numerator and the derivative of the denominator separately: [ \lim_{x \to 0} \frac{e^{3x} - e^{5x}}{x} = \lim_{x \to 0} \frac{d}{dx}(e^{3x}) - \frac{d}{dx}(e^{5x}) = \lim_{x \to 0} \frac{3e^{3x} - 5e^{5x}}{1} ]

Now, evaluate the limit again: [ \lim_{x \to 0} \frac{3e^{3x} - 5e^{5x}}{1} = \frac{3e^{3(0)} - 5e^{5(0)}}{1} = \frac{3e^0 - 5e^0}{1} = \frac{3 - 5}{1} = -2 ]

Therefore, the limit of ( \frac{e^{3x} - e^{5x}}{x} ) as ( x ) approaches 0 is ( -2 ).

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Use l'Hospital's Rule.

We can apply l"Hospital's rule to try to find such a limit.

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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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