lim as x approaches 0
(e^4x-1)/sin(2x)
I know you use L-hopitals rule and I know the answer is 2 but I keep getting 1. Can someone help
lim as x approaches 0
(e^4x-1)/sin(2x)
assuming you mean:
(e^(4x) - 1)/sin(2x)
= lim (4 e^(4x))/(2cos(2x)) as x ---> 0
= 4e^0 / 2cos0
= 4(1)/2 = 2
Thank you that makes sense!
Sure, I can help you with that!
To evaluate the limit as x approaches 0 of the expression (e^4x - 1) / sin(2x), you're on the right track by using L'Hôpital's rule. L'Hôpital's rule allows you to take the derivative of both the numerator and the denominator separately until you reach an indeterminate form (such as 0/0 or ∞/∞).
Let's go through the steps together:
1. Take the derivative of the numerator:
Derivative of (e^4x - 1) with respect to x:
lim as x approaches 0 of 4e^4x = 4.
2. Take the derivative of the denominator:
Derivative of sin(2x) with respect to x:
lim as x approaches 0 of 2cos(2x) = 2.
Now, we have (4)/(2) = 2, which is the correct answer.
If you obtained a different result, it's possible that there was a mistake in taking the derivatives or in the arithmetic calculations. Double-check your work to ensure you took the derivatives correctly and performed the calculations accurately.
Remember to simplify the original expression before applying L'Hôpital's rule, as the rule only applies to indeterminate forms. In this case, the limit was in indeterminate form (0/0) because both the numerator and the denominator approached 0 as x approaches 0.