Find the following limits using L'Hospitals rule.?
A is in the form infinity over infinity; take the derivatives of the numerator and denominator: 4e^4x / 3 As x goes to infinity, the numerator goes to infinity while the denominator stays at 3, so your answer is infinity. B is in the form of 0 * negative infinity; divide the terms and give one term a negative exponent: (ln x)/(x^-2) Now it's in the form of negative infinity over infinity; differentiate the numerator and denominator: (1/x) / (-2x^-3) As x goes to 0, that goes to infinity over zero, so the whole thing is going to infinity (when you multiply something by infinity or divide by zero, it goes to infinity; you're doing both here, so they both go to infinity). C is in the form of 1^(negative infinity); take e^[ln(the original problem)]: e^{ln[(1+3/x)^-x]} Use log laws to bring down the -x: e^{-x * ln[(1+3/x)]} As x goes to infinity, you have e to the power of (infinity times zero). Raise one of those terms to the power of negative one and divide them: e^{ln[(1+3/x)] / x^(-1)} As x goes to infinity, that's e to the power of infinity over infinity; go from there.
Find the Following Limit using L'Hopital's Rule if Appropriate?
L = lim ( x → ∞ ) [ ( arctan x ) / ( (1/x) - 8 ). This limit is not in the indeterminate form 0/0 or ∞/∞. Hence, we cannot use L' Hopital's Rule even if it were necessary which, in this case, isn't. Direct substitution gives L = ( arctan ∞ ) / ( 0 - 8 ) = ( π/2 ) / ( - 8 ) = -π/16 ....... Ans. ......................................... Happy To Help ! .........................................
Find the following limit without using a graphing calculator or making a table?
6x, use something called L'Hopital's Rule. if you plug the limit h into the equation you get 0/0. if you get 0/0 you can use L'Hopital's Rule. This rule says to take the derivative of the top and derivative separately. so take the derivative of 6xh-3h^2 with respect to h and then take the derivative of h with respect to h. this gives you 6x-6h/1. plug 0 into this and it gives you 6x. if you're not in calculus then i forget how to do it.
Find the following limit if it exists: lim ln(x+3) as X approaches -3 from the left?
The limit doesn't exist when approaching from the left as the function is not defined for (∞,3]. If you approach from the right then x+3 will approach 0 from the right. And ln will approach -∞ as x+3 gets closer and closer to 0.
How would one find the following limit (no L'Hopital's Rule, Taylor Series, etc.)?
lim (x-->0) (x^2 - sin^2(x)) / (x^2 sin^2(x)) = lim (x-->0) (1 - (sin(x)/x)^2) / (1 - cos^2(x)) = lim (x-->0) [ (1 - (sin(x)/x)) / (1 - cos(x)) ] * [ (1 + (sin(x)/x)) / (1 + cos x) ] = A * B, where lim (x-->0) [ 1 - (sin(x)/x)) / (1 - cos(x)) ] = A and lim (x-->0) [ (1 + (sin(x)/x)) / (1 + cos x) ] = B With the knowledge that lim (x-->0) sin(x)/x = 1, it is easy to evaluate that B = 1. Thus the limit we are trying to evaluate is A. lim (x-->0) (1 - (sin(x)/x)) / (1 - cos(x)) = lim (x-->0) (1 - (sin(x)/x)) / [ x^2 (1 - cos(x)) / x^2 ] = A = C / D, where lim (x-->0) [ (1 - (sin(x)/x)) / x^2 ] = C and lim (x-->0) [ (1 - cos(x)) / x^2 ] = D To evaluate D, let x = 2a. (Note that 2a --> 0 is the same as a --> 0) lim (x-->0) [ (1 - cos(x)) / x^2 ] = lim (a-->0) [ (1 - cos(2a)) / (4a^2) ] = lim (a-->0) [ (1 - (1 - 2sin^2(a)) / (4a^2) ] = lim (a-->0) [ (2sin^2(a)) / (4a^2) ] = lim (a-->0) [ ((sin(a)/a)^2) / 2 ] Since lim (a-->0) sin(a)/a = 1, D = 1/2. Hence A = 2C. Now we evaluate C. lim (x-->0) (1 - (sin(x)/x)) / x^2 = lim (x-->0) (x - sin(x)) / x^3 Let x = 3b. (Once again, 3b --> 0 is the same as b -->0) lim (x-->0) (x - sin(x)) / x^3 = lim (b-->0) (3b - sin(3b)) / (27b^3) = lim (b-->0) (3b - (3sin(b) - 4sin^3(b)) / (27b^3) = lim (b-->0) (b - sin(b)) / (9b^3) + (4/27)(sin(b)/b)^3 = C = E + F, where lim (b-->0) [ (b - sin(b)) / (9b^3) ] = E and lim (x-->0) (4/27)(sin(b)/b)^3 = F For the reason mentioned many times before, F = 4/27. Note that E = C/9. Thus, C/9 + 4/27 = C. 8C/9 = 4/27. C = 1/6. A = 2C, therefore A = 1/3 and hence lim (x-->0) (x^2 - sin^2(x))/(x^2 sin^2(x)) = 1/3. The source below has been of great help to the production of this answer.
Compute the following limits using l'H\^opital's rule if appropriate.?
1)5sin(5x)/4sin(4x)-->25cos((5x)/16cos(4... =25/16 2)5^xln5-4^xln4/2x=5ln5-4ln4/2