Composite Indeterminate Forms

Composite Forms of Limits

In some limits, the given expression consists of multiple terms that interact in a way that leads to composite indeterminate forms. These cases are more complicated than standard indeterminate forms because different parts of the expression may tend to infinity, zero, or other undefined values at different rates. Such limits require careful simplification to determine the actual result.

Consider the example:

\[ \lim_{x \to \infty} \left( x - \frac{x^3 + 1}{x^2 + x + 1} \right) ^{\left( \frac{x^2}{x+1} \right)}. \]

At first glance, each term inside the limit appears to grow infinitely large. The first term contains a difference where one part tends to infinity and the other is a fraction with an indeterminate form \( \frac{\infty}{\infty} \). This creates an overall form:

\[ \left( \infty - \frac{\infty}{\infty} \right) ^ {\left( \frac{\infty}{\infty} \right)}. \]

Such cases require identifying dominant terms and simplifying the expression properly, as direct substitution does not provide a clear answer. Understanding how different terms behave as \( x \to \infty \) allows us to determine the actual limit, avoiding incorrect conclusions based on raw infinity operations.

To solve, we first determine each known form. In the given expression, we must first evaluate

\[ \lim_{x \to \infty} \frac{x^3 + 1}{x^2 + x + 1} \]

which is clearly of the form \( \frac{\infty}{\infty} \). Since the degree of the numerator is higher than that of the denominator, we immediately conclude that this limit is \( \infty \).

Next, we analyze the exponent, which also contains an indeterminate form:

\[ \lim_{x \to \infty} \frac{x^2}{x+1}. \]

Again, since the degree of the numerator is greater than the degree of the denominator, this limit is also \( \infty \).

Thus, the original expression reduces to the form:

\[ (\infty - \infty)^\infty. \]

Now, in the base, we encounter another indeterminate form of the type \( \infty - \infty \). To resolve this, we must first evaluate:

\[ \lim_{x \to \infty} \left( x - \frac{x^3 + 1}{x^2 + x + 1} \right). \]

To solve the \( \infty - \infty \) form, we know that we need to combine both terms. In this limit, we take the LCM, which gives us:

\[ \lim_{x \to \infty} \frac{x^2 + x - 1}{x^2 + x + 1}. \]

This is again an \( \frac{\infty}{\infty} \) form. Since the degree of the numerator is equal to the degree of the denominator, the limit is given by the ratio of their leading coefficients. The leading coefficient of both the numerator and denominator is 1, so we get:

\[ \lim_{x \to \infty} \frac{x^2 + x - 1}{x^2 + x + 1} = 1. \]

Ultimately, the base in the original problem is approaching 1, and the exponent is approaching \( \infty \). This gives us the form:

\[ 1^\infty. \]

Now, we return to the original problem, which has the indeterminate form \( 1^\infty \).

We have the given limit:

\[ \lim_{x \to \infty} \left( x - \frac{x^3 + 1}{x^2 + x + 1} \right) ^{\left( \frac{x^2}{x+1} \right)} \]

which we have already determined to be of the indeterminate form \( 1^\infty \). To resolve this, we use the standard result:

\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) (f(x) - 1)}. \]

From our previous calculation, the base approaches:

\[ \lim_{x \to \infty} \frac{x^2 + x - 1}{x^2 + x + 1} = 1. \]

Thus, applying the logarithm transformation,

\[ \lim_{x \to \infty} e^{\left( \frac{x^2}{x+1} \right) \left( \frac{x^2 + x - 1}{x^2 + x + 1} - 1 \right)}. \]

Since:

\[ \frac{x^2 + x - 1}{x^2 + x + 1} - 1 = \frac{(x^2 + x - 1) - (x^2 + x + 1)}{x^2 + x + 1} = \frac{-2}{x^2 + x + 1}, \]

the exponent simplifies to:

\[ \left( \frac{x^2}{x+1} \right) \left( \frac{-2}{x^2 + x + 1} \right). \]

Since the degree of the numerator in this product is less than the degree of the denominator, the entire expression tends to 0 as \( x \to \infty \), giving:

\[ e^0 = 1. \]

Thus, the final result is:

\[ \lim_{x \to \infty} \left( x - \frac{x^3 + 1}{x^2 + x + 1} \right) ^{\left( \frac{x^2}{x+1} \right)} = 1. \]