Is this supposed to be a definite integral? If so, is this the problem?Shannabel said:Homework Statement
evaluate lim(x->0) (tan^8(t))dt(between 0 and sin^2x)
How did you get this?Shannabel said:Homework Equations
The Attempt at a Solution
[tan^8(sin^2(x))]/sin^18(x)
Shannabel said:my book says to use l'hospital's rule, so i continued with
[8tan^7(sin^2x)*sec^2(sin^2(X))*2sinxcosx]
but my book says i should just have
[tan^8(sin^2x)*2sinxcosx]/18sin^17(x)cos(x)
why did they only partially differentiate tan^8(sin^2(x))?
Mark44 said:Is this supposed to be a definite integral? If so, is this the problem?
[tex]\lim_{x \to 0} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
You can see the LaTeX I used by clicking on the integral.
How did you get this?
Shannabel said:no, it should be
[tex]\lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
Shannabel said:no, it should be
[tex]\lim_{x \to 0} \frac{1}{sin^{18}(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
Shannabel said:no, it should be
[tex]\lim_{x \to0} 1/{sin^{18}(x)} \int_0^{sin^2(x)} tan^8(t)~dt[/tex]
Mark44 said:OK, that makes more sense. L'Hopital's is applicable here.
In answer to your questions, they didn't "partially differentiate" the expression in the integrand - they used the chain rule.
You know (I think) this part of the Fund. Thm. of Calculus involving derivatives and definite integrals.
[tex]\frac{d}{dx} \int_a^x f(t)~dt~= f(x)[/tex]
If the limit of integration contains a function of x, then you need to use the chain rule.
[tex]\frac{d}{dx} \int_a^{g(x)} f(t)~dt~= \frac{d}{du} \int_a^{u} f(t)~dt * \frac{du}{dx}~= f(u) * du/dx[/tex]
L'Hospital's rule with trig functions is a mathematical rule used to evaluate limits involving trigonometric functions. It states that if the limit of a quotient of two functions is indeterminate (i.e. it has the form 0/0 or ∞/∞), then the limit of the quotient of the derivatives of those functions will be the same as the limit of the original quotient.
To apply L'Hospital's rule with trig functions, you first need to determine if the limit is indeterminate. If it is, then you can take the derivative of the numerator and denominator separately. After taking the derivatives, you can evaluate the new quotient, which should give you the same limit as the original one.
One example of using L'Hospital's rule with trig functions is evaluating the limit of sin(x)/x as x approaches 0. Another example is evaluating the limit of (1-cos(x))/x^2 as x approaches 0. Both of these examples result in an indeterminate form, and L'Hospital's rule can be applied to find the limit.
Yes, L'Hospital's rule can be used with any type of trigonometric function, including sine, cosine, tangent, cotangent, secant, and cosecant. However, the limit must still be indeterminate in order for the rule to be applied.
Yes, there are some limitations to L'Hospital's rule with trig functions. It can only be used when the limit is indeterminate, and it may not always give the correct result if the limit does not exist. It is also important to check for other possible methods of evaluating the limit before applying L'Hospital's rule.