Evaluating Indeterminate Limits Using Taylor Series

In certain cases, the Taylor series provides an effective technique for resolving indeterminate limits.

A Worked Example

Consider the following limit, which exhibits an indeterminate form of type ∞ - ∞:

$$ \lim_{x \rightarrow 0} \left( \frac{1}{x^2} - \frac{1}{x \sin x} \right) = \infty - \infty $$

To tackle this expression, we approximate the sine function using its first-order Taylor expansion centered at $ x = 0 $:

$$ \sin x = x - \frac{x^3}{6} + o(x^4) $$

Before substitution, we rewrite the expression as a single rational function to simplify manipulation:

$$ \frac{1}{x^2} - \frac{1}{x \sin x} = \frac{x \sin x - x^2}{x^3 \sin x} = \frac{\sin x - x}{x^2 \sin x} $$

Now substitute the Taylor approximation:

$$ \frac{ \left( x - \frac{x^3}{6} + o(x^4) \right) - x }{ x^2 \left( x - \frac{x^3}{6} + o(x^4) \right) } = \frac{ -\frac{x^3}{6} + o(x^4) }{ x^3 - \frac{x^5}{6} + o(x^6) } $$

We now simplify by dividing numerator and denominator by $ x^3 $:

$$ \frac{ -\frac{1}{6} + \frac{o(x^4)}{x^3} }{ 1 - \frac{x^2}{6} + \frac{o(x^6)}{x^3} } $$

Taking the limit as $ x \to 0 $:

$$ \lim_{x \rightarrow 0} \frac{ -\frac{1}{6} + \frac{o(x^4)}{x^3} }{ 1 - \frac{x^2}{6} + \frac{o(x^6)}{x^3} } $$

We split the expression into the ratio of two limits:

$$ \frac{ \lim_{x \to 0} \left( -\frac{1}{6} + \frac{o(x^4)}{x^3} \right) }{ \lim_{x \to 0} \left( 1 - \frac{x^2}{6} + \frac{o(x^6)}{x^3} \right) } $$

Constants can be pulled out of the limits:

$$ \frac{ -\frac{1}{6} + \lim_{x \to 0} \frac{o(x^4)}{x^3} }{ 1 - \lim_{x \to 0} \frac{x^2}{6} + \lim_{x \to 0} \frac{o(x^6)}{x^3} } $$

Since $ \lim_{x \to 0} \frac{x^2}{6} = 0 $, we simplify the denominator:

$$ \frac{ -\frac{1}{6} + \lim_{x \to 0} \frac{o(x^4)}{x^3} }{ 1 + \lim_{x \to 0} \frac{o(x^6)}{x^3} } $$

To evaluate the little‑o terms, we factor out powers of $ x $:

$$ \frac{ -\frac{1}{6} + \lim_{x \to 0} \left( x \cdot \frac{o(x^4)}{x^4} \right) }{ 1 + x^3 \cdot \lim_{x \to 0} \frac{o(x^6)}{x^6} } $$

Since $ o(x^4) $ is a higher-order infinitesimal than $ x^4 $, the term $ \frac{o(x^4)}{x^4} \to 0 $ as $ x \to 0 $. Likewise, $ \frac{o(x^6)}{x^6} \to 0 $:

$$ \frac{ -\frac{1}{6} + 0 }{ 1 + 0 } = -\frac{1}{6} $$

Therefore, the value of the limit is:

$$ \boxed{ -\frac{1}{6} } $$

This result shows how Taylor expansions can simplify and resolve seemingly intractable indeterminate forms.

And so on.