Evaluating Limits Using L'H?pital's Rule: A Comprehensive Guide
In calculus, evaluating limits can be a challenging task. One common scenario involves evaluating limits of the form 0/0, where both the numerator and the denominator approach zero as x approaches a certain value. In such cases, L'H?pital's Rule is a powerful tool that can simplify the process. This article will delve into the details of how to use L'H?pital's Rule to evaluate such limits, with a focus on a specific case involving cube roots and square roots.
Introduction to L'H?pital's Rule
L'H?pital's Rule is a theorem that allows us to evaluate limits of the form 0/0 or ∞/∞. The rule states that if the limit of the function f(x)/g(x) as x approaches a certain value (say 1) is of the form 0/0 or ∞/∞, then the limit can be evaluated by taking the limit of the derivatives of the numerator and the denominator, provided the limit exists. In this case, L lim (x→a) [f'(x)/g'(x)].
Evaluating the Limit
Let's consider the limit limx→1 ( (x1/3-1)/(x1/2-1) ). At first glance, it may seem challenging due to the indeterminate form 0/0. However, we can apply L'H?pital's Rule to simplify the expression.
Step 1: Differentiate the Numerator and Denominator
Numerator: sqrt[3]{x} - 1Differentiate the numerator: diff sqrt[3]{x} 1/3 x-2/3 1/3sqrt[3]{x2}Denominator: sqrt{x} - 1Differentiate the denominator: diff sqrt{x} 1/2 x-1/2 1/2sqrt{x}Step 2: Apply L'H?pital's Rule
By applying L'H?pital's Rule, we get:
[ lim_{x to 1} frac{sqrt[3]{x}-1}{sqrt{x}-1} lim_{x to 1} frac{frac{1}{3}sqrt[3]{x^2}}{frac{1}{2}sqrt{x}} ]Step 3: Simplify the Limit
Simplifying the expression, we obtain:
[ lim_{x to 1} frac{frac{1}{3}sqrt[3]{x^2}}{frac{1}{2}sqrt{x}} lim_{x to 1} frac{2sqrt{x}}{3sqrt[3]{x^2}} ]Step 4: Evaluate the Limit
Finally, we substitute x 1:
[ lim_{x to 1} frac{2sqrt{x}}{3sqrt[3]{x^2}} frac{2sqrt{1}}{3sqrt[3]{1^2}} frac{2}{3} ]Thus, the value of the limit is boxed{2/3}.
Alternative Methods
There are other methods to evaluate limits, such as using the Taylor expansion. For example, consider the limit with the numerator and denominator of the form of a difference of powers. Let's take the limit:
[ lim_{x to 1} frac{x^{1/6} - 1}{x^{1/3} - 1} ]Step 1: Apply the Disuity Rule
The key idea is to express the difference as a product and then simplify:
[ frac{x^{1/6} - 1}{x^{1/3} - 1} frac{(x^{1/6}) - 1}{x^{1/6}} cdot frac{x^{1/6}}{x^{1/3} - 1} frac{x^{1/6} - 1}{x^{1/6}} cdot frac{1}{x^{1/6}} cdot frac{x^{1/6}}{x^{1/3} - 1} ]Substituting back and simplifying:
[ lim_{x to 1} frac{1}{3} cdot frac{(x^{1/6}) - 1}{x^{1/6}} cdot frac{2}{x^{1/3} - 1} frac{1^{1/3} cdot 1^{1/6} cdot 2}{1^{1/6} cdot 3} frac{3}{2} ]Conclusion
In summary, L'H?pital's Rule is a powerful method for evaluating limits of the form 0/0. While alternative methods like Taylor expansion can be equally useful, L'H?pital's Rule often simplifies the process. The key steps involve differentiating the numerator and denominator, applying the rule, and then simplifying or evaluating the resulting expression.
Remember, the golden rule for evaluating limits at a point is to consider the lowest order terms in the Taylor expansion. This can often provide a straightforward way to evaluate the limit.