Solving Indeterminate Forms Using L'H?pital's Rule and Substitution
The concept of indeterminate forms is an important aspect of calculus, particularly when dealing with limits. In this discussion, we will delve into the solution of such forms using L'H?pital's Rule and substitution techniques. Specifically, we will explore the indeterminate form (frac{0}{0}) and analyze a particular problem to illustrate these methods.
Problem Statement and Initial Setup
We are given the following limit:
(lim_{x to -2} frac{frac{1}{x} - frac{1}{2}}{x^3 - 8})
This limit presents an indeterminate form (frac{0}{0}) when we substitute (x -2), as both the numerator and the denominator approach zero. To resolve this, we can use L'H?pital's Rule or substitution to find the limit.
L'H?pital's Rule Approach
Firstly, let's apply L'H?pital's Rule, which states that if the limit of the ratio of two functions as (x) approaches a certain value is an indeterminate form (frac{0}{0}) or (frac{infty}{infty}), then the limit of the ratio of their derivatives is the same.
[ lim_{x to -2} frac{frac{1}{x} - frac{1}{2}}{x^3 - 8} lim_{x to -2} frac{frac{d}{dx}left(frac{1}{x} - frac{1}{2}right)}{frac{d}{dx}(x^3 - 8)} ]
First, compute the derivatives of the numerator and the denominator:
[ frac{d}{dx}left(frac{1}{x} - frac{1}{2}right) -frac{1}{x^2} ]
and
[ frac{d}{dx}(x^3 - 8) 3x^2 ]
Substituting these derivatives back into the limit:
[ lim_{x to -2} frac{-frac{1}{x^2}}{3x^2} lim_{x to -2} -frac{1}{3x^4} ]
Substitute (x -2) into the expression:
[ -frac{1}{3 times 16} -frac{1}{48} ]
Therefore, the limit is:
(lim_{x to -2} frac{frac{1}{x} - frac{1}{2}}{x^3 - 8} -frac{1}{48})
Substitution Approach
Another method to solve this problem is by using a substitution technique. Let's rewrite the original limit using a substitution to simplify the expression:
Let (y x 2), hence as (x to -2), (y to 0). The expression can be transformed as:
[ lim_{x to -2} frac{frac{1}{x} - frac{1}{2}}{x^3 - 8} lim_{y to 0} frac{frac{1}{y-2} - frac{1}{2}}{(y-2)^3 - 8} ]
Simplify the expression inside the limit:
[ lim_{y to 0} frac{frac{1}{y-2} - frac{1}{2}}{(y-2)^3 - 8} lim_{y to 0} frac{frac{2 - (y-2)}{2(y-2)}}{y^3 - 6y 12} ]
Further simplifying:
[ lim_{y to 0} frac{frac{4-y}{2(y-2)}}{(y-2)left(y^2 6right)} lim_{y to 0} frac{4-y}{2y(y-2)^2(y 6)} ]
Cancel out the ((y-2)) terms:
[ lim_{y to 0} frac{4-y}{2y(y 6)(y-2)} lim_{y to 0} frac{1}{2yleft(y 6right)} ]
Substitute (y 0) into the expression:
[ frac{1}{2 times 6} -frac{1}{48} ]
Therefore, the limit using the substitution approach is also:
(lim_{x to -2} frac{frac{1}{x} - frac{1}{2}}{x^3 - 8} -frac{1}{48})
Conclusion
Both methods, L'H?pital's Rule and substitution, provide the same result, (-frac{1}{48}). The problem highlights the importance of dealing with indeterminate forms in calculus and demonstrates two effective techniques for resolving them.