lzkelley said:well, we can see that the answer is going to be one, because in the limit that x --> inf, x = x+1 therefore x/x+1 = 1, so 1^anything even inf is still 1.
Now to prove that more rigorously, you break it up into numerator and denominator, f(x)/g(x) let's say, where f(x) = x^7x and g(x) = (x+1)^7x l'hopital's rule says that if both f(x) and g(x) are divergent, then x--> inf f(x)/g(x) = x-->inf f'(x) / g'(x) , so you take the derivative of the top and bottom separately... you're doing to have to do this until its not divergent anymore... or 8 times? i guess?
i see what you're going for with the exp(7xln...) but i think that's probably harder / more work. Good luck
L'Hospital's rule is a mathematical theorem that allows for the evaluation of certain limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of a quotient of two functions at a certain point is indeterminate, then the limit of the quotient of the derivatives of the two functions at that point will be the same.
L'Hospital's rule should only be used when the limit of a quotient of two functions at a certain point is indeterminate. It should not be used for limits that do not have an indeterminate form, as it may give incorrect results.
The conditions for L'Hospital's rule to apply are that the limit being evaluated must have an indeterminate form (0/0 or ∞/∞), the functions involved must be differentiable in a neighborhood of the point in question, and the limit of the quotient of the derivatives must exist.
Yes, there are limitations to L'Hospital's rule. It cannot be applied to limits involving certain types of indeterminate forms, such as ∞ - ∞ or 0 * ∞, and it cannot be used to evaluate limits at points where the functions involved are not differentiable.
No, L'Hospital's rule cannot be used to solve all limits. It is only applicable to limits involving indeterminate forms, and there are many limits that do not have an indeterminate form and therefore cannot be evaluated using L'Hospital's rule.