i need help with proving five steps of l hospitals rule for (ex^2-1)/x
0: (e^x^2-1)/x
1: 2xe^x/1
2: 2e^x(x+1)/0
??? Not sure what the five steps are. Obviously it's not 5 iterations.
just carry tha too yA crazy kiddo!
To prove the five steps of l'Hôpital's Rule for the function (ex^2 - 1) / x, follow these steps:
Step 1: Check the limiting form
To apply l'Hôpital's Rule, we need to check if the function has an indeterminate form at the limit point. In this case, when x approaches 0, the function becomes (e(0)^2 - 1) / 0 = (e^0 - 1) / 0 = (1 - 1) / 0, which is a 0/0 indeterminate form.
Step 2: Differentiate the numerator and denominator
Differentiate the numerator, which is ex^2 - 1, and the denominator, which is x, separately.
Derivative of numerator: (d/dx) (ex^2 - 1) = 2x * ex^2 = 2xex^2 = 2ex^3.
Derivative of denominator: (d/dx) (x) = 1.
Step 3: Evaluate the limit of the derivatives
Take the limit as x approaches 0 of the derivatives obtained in Step 2.
Limit of the derivative of numerator: lim(x->0) 2ex^3 = 2e(0)^3 = 2e(0) = 2e.
Limit of the derivative of denominator: lim(x->0) 1 = 1.
Step 4: Apply l'Hôpital's Rule
If the limit in Step 3 exists or is ±∞, then we can apply l'Hôpital's Rule by taking the derivative of the numerator and denominator again.
Derivative of numerator: (d/dx) (2ex^3) = 6x^2 * ex^3 = 6x^2ex^3 = 6ex^5.
Derivative of denominator: (d/dx) (1) = 0.
Step 5: Evaluate the final limit
Take the limit as x approaches 0 of the second set of derivatives obtained in Step 4.
Limit of the second derivative of numerator: lim(x->0) 6ex^5 = 6e(0)^5 = 6e(0) = 6e.
Limit of the second derivative of denominator: lim(x->0) 0 = 0.
If the limit in Step 5 exists or is ±∞, then this is the final result. In this case, the limit of the function as x approaches 0 is 6e.
Therefore, by following these five steps, we have successfully proven l'Hôpital's Rule for the function (ex^2 - 1) / x.