Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0 (e^9x−1−9x)/(x^2)
To find the limit of the given function, we can apply L'Hospital's Rule. This rule allows us to evaluate limits involving indeterminate forms such as 0/0 or ∞/∞. In this case, we have a form of 0/0, thus we can proceed with L'Hospital's Rule.
L'Hospital's Rule states that if the limit of the ratio of two functions, f(x) and g(x), as x approaches a certain value (in this case, 0), is of an indeterminate form, then the limit of the ratio is equal to the limit of the ratio of their derivatives.
Let's differentiate the numerator and the denominator separately:
Numerator: (e^(9x) - 1 - 9x)'
Using the chain rule, the derivative of e^(9x) is 9e^(9x). The derivative of -1 is 0, and the derivative of 9x is 9. Hence, the numerator's derivative is 9e^(9x) - 9.
Denominator: (x^2)'
Differentiating x^2 with respect to x, we get 2x.
Now, let's rewrite the limit using the derivatives:
lim x→0 (9e^(9x) - 9)/(2x)
Now we can evaluate the limit directly by substituting x = 0 into the expression:
lim x→0 (9e^(9x) - 9)/(2x) = (9e^(0) - 9)/(2(0)) = (9 - 9)/(0) = 0/0
We can see that after evaluating the limit using L'Hospital's Rule once, we reached another indeterminate form: 0/0. Therefore, let's apply L'Hospital's Rule again.
Differentiating the numerator and the denominator:
Numerator: (9e^(9x) - 9)'
The derivative of 9e^(9x) is 9 * 9e^(9x), which simplifies to 81e^(9x).
Denominator: (2x)'
The derivative of 2x is 2.
Now, we can rewrite the limit again using the new derivatives:
lim x→0 (81e^(9x))/(2)
Evaluating the limit:
lim x→0 (81e^(9x))/(2) = (81e^(9 * 0))/(2) = 81e^0/2 = 81/2
Therefore, the limit of the given function, as x approaches 0, is 81/2.