Evaluating Limits Involving Exponential Functions and Indeterminate Forms
In mathematics, particularly when dealing with limits, we often encounter expressions that seem to fall into indeterminate forms. One such common form is the expression [lim_{x to a} left( frac{f(x) - 1}{x - 3} right)^{x - 3}]. This article will demonstrate how to evaluate the specific limit as [x to 0] for the expression [left( frac{x - 1}{x - 3} right)^{x - 3}].
Understanding Indeterminate Forms
When evaluating limits, we sometimes encounter indeterminate forms such as [frac{0}{0}], [frac{infty}{infty}], or [0^0]. These forms are called indeterminate because they cannot be directly evaluated and require further analysis. However, for the given expression, these forms do not apply as we shall see.
Evaluating the Limit Directly
Sometimes, the limit can be evaluated directly by substitution. In the case of the expression [left(frac{x - 1}{x - 3}right)^{x - 3}], let's substitute [x 0].
[left(frac{0 - 1}{0 - 3}right)^{0 - 3} left( frac{-1}{-3} right)^{-3} frac{1}{3}^{-3} 27]
Therefore, the limit is evaluated as [27]. This is one of the possible forms of evaluating the limit.
Using L'H?pital's Rule and Exponential Form
Another method to evaluate such limits is by converting the expression into an exponential form. We can use the property [lim_{x to a} f(x)^{g(x)} e^{lim_{x to a} g(x) ln f(x)}] to transform the given limit.
[lim_{x to 0} left(frac{x - 1}{x - 3}right)^{x - 3} lim_{x to 0} e^{(x - 3) ln left(frac{x - 1}{x - 3}right)}]
Next, we simplify the expression inside the exponent:
[lim_{x to 0} (x - 3) ln left(frac{x - 1}{x - 3}right) lim_{x to 0} (x - 3) left(ln left(frac{x - 1}{x - 3}right) - ln(1)right) -3 ln left(frac{-2}{-3}right) -3 ln left(frac{2}{3}right)]
Finally, we evaluate the limit as:
[lim_{x to 0} e^{(x - 3) ln left(frac{x - 1}{x - 3}right)} e^{-3 ln left(frac{2}{3}right)} left(frac{2}{3}right)^{-3} 27]
Thus, the limit evaluates to [27].
Conclusion
In conclusion, the limit of the expression [left(frac{x - 1}{x - 3}right)^{x - 3}] as [x to 0] is [27]. This evaluation can be done both directly by substitution and through the use of exponential forms and L'H?pital's Rule. Understanding these techniques can be crucial in solving similar limit problems involving indeterminate forms.