Question

Explain how to convert a limit of the form 0•[infinity] to a limit of the form 0/0 or [infinity]/[infinity]

A. If lim X-a f(x)g(x) has the indeterminate form 0•[infinity], then lim X-a f(x)/g(x) has the indeterminate form 0/0 and lim X-a g(x)/f(x) has the indeterminate form [infinity]/[infinity].
B. If lim X-a f(x)g(x) has the indeterminate form 0•[infinity], then lim X-a f(x)/g(x) has the indeterminate form 0/0 or [infinity]/[infinity].
C. If lim X-a f(x)g(x) has the indeterminate form 0•[infinity], then lim X-a f(x) / 1/g(x) has the indeterminate form 0/0 or [infinity]/[infinity], provided lim X-a 1/g(x) exists or is infinite.
D. If lim X-a f(x)g(x) has the indeterminate form 0•[infinity], then lim X-a g(x) / 1/f(x) has the indeterminate form 0/0, but not [infinity]/[infinity], and lim X-a 1/g(x) has the indeterminate form [infinity]/[infinity], but not 0/0, provided lim 1/f(x) exists or is infinite.

Answer 1

To address the indeterminate form 0·[∞], algebraically manipulate it into the form 0/0 or [∞]/[∞] by taking the reciprocal of one function, enabling the use of L'Hôpital's Rule.

Step-by-step explanation:

To convert a limit of the form 0·[∞] (zero times infinity) to a limit of the form 0/0 or [∞]/[∞], we use algebraic manipulation. The correct transformation is described by option C, which states: If lim X-a f(x)g(x) has the indeterminate form 0·[∞], then lim X-a f(x) / 1/g(x) has the indeterminate form 0/0 or [∞]/[∞], given that lim X-a 1/g(x) exists or is infinite. To apply this transformation, we define one of the functions as 1/another function, essentially flipping one function to its reciprocal to create a quotient instead of a product. This algebraic manipulation allows us to use L'Hôpital's Rule, which applies to 0/0 or [∞]/[∞] cases but not to 0·[∞].

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