Limits: indeterminate forms

The question of solving limits is quite extensive, since there are dozens of methods for solving limits of various kinds. When we have a limit, such as $\lim\limits_{x \to a} f(x)$, at first we just try to substitute $a$ into the function $f(x)$. But we can get expressions like the following:

$$\dfrac{0}{0}, \ \dfrac{\infty}{\infty}, \ 1^{\infty}, \ 0 \cdot \infty, \ \infty-\infty, \ \infty^0, \ 0^0$$These are indeterminate forms. We can't interpret this result: is it zero, is it undefined or is it something else? Fortunately, there are several rules we can apply in such cases.

Limits with $\dfrac{\infty}{\infty}$ indeterminate form

Now we consider limits when $x \to \infty$ and the function is a fraction and contains polynomials in its numerator and denominator .

Suppose we need to calculate $$\lim\limits_{x \to \infty} \dfrac{8x^3 + 2x^2 + x + 3}{4x^4+3x^2 + 5}$$If we try to substitute infinity instead of $x$, we will get $\dfrac{\infty}{\infty}$ indeterminate form. One would think that the answer is $\infty$, but in general case this is not a correct answer. How do we solve limits of this type? First, we look at the numerator and find the highest degree of $x$: it is $x^3$. Secondly, we look at the denominator and also find the highest degree of $x$: it is $x^4$. Then we select the highest degree between the numerator and denominator: in this example it is $x^4$. And then we divide the numerator and denominator by $x$ to the highest degree:

$$\lim\limits_{x \to \infty} \dfrac{\dfrac{8x^3}{x^4} + \dfrac{2x^2}{x^4} + \dfrac{x}{x^4} + \dfrac{3}{x^4} }{\dfrac{4x^4}{x^4} +\dfrac{3x^2}{x^4} +\dfrac{5}{x^4} } =\lim\limits_{x \to \infty} \dfrac{\dfrac{8}{x} + \dfrac{2}{x^2} + \dfrac{1}{x^3}+\dfrac{3}{x^4}}{4 + \dfrac{3}{x^2} + \dfrac{5}{x^4}} = \\ \ \\ \dfrac{\lim\limits_{x \to \infty}\dfrac{8}{x} + \lim\limits_{x \to \infty}\dfrac{2}{x^2} + \lim\limits_{x \to \infty}\dfrac{1}{x^3}+\lim\limits_{x \to \infty}\dfrac{3}{x^4}}{\lim\limits_{x \to \infty}4 + \lim\limits_{x \to \infty}\dfrac{3}{x^2} + \lim\limits_{x \to \infty}\dfrac{5}{x^4}} = \dfrac{0}{4} = 0$$So, the solution method is as follows: in order to reveal the indeterminate form, we need to divide the numerator and the denominator by the highest degree of $x$.

Limits with $\dfrac{0}{0}$ indeterminate form

This group of limits is somewhat similar to the limits we have just considered: in the numerator and denominator there are polynomials, but now $x \to a$, where $a$ is a finite number.

Consider the following example:

$$\lim\limits_{x \to -2} \dfrac{2x^2 + 7x + 6}{x+2}$$If we try to substitute $-2$ instead of $x$, we will get the $\dfrac{0}{0}$ indeterminate form. To disclose this indeterminate form, we need to factor the numerator and denominator. We notice that $$2x^2 + 7x + 6 = (x+2)(2x+3)$$Hence,

$$\lim\limits_{x \to -2} \dfrac{2x^2 + 7x + 6}{x+2} = \lim\limits_{x \to -2} \dfrac{(x+2)(2x+3)}{x+2} = \lim\limits_{x \to -2} (2x+3) = -1$$

Limits with $\infty - \infty$ indeterminate form

Suppose we need to find

$$\lim \limits_{x \to 5} \left( \dfrac{10}{x^2-25} - \dfrac{1}{x-5}\right)$$If we try to substitute $5$ instead of $x$, we will get the $\infty - \infty$ indeterminate form. In this case, the general solution algorithm is as follows: we need to bring the expression to a common denominator and then try to reduce something:

$$\lim \limits_{x \to 5} \left( \dfrac{10}{x^2-25} - \dfrac{1}{x-5}\right) = \lim \limits_{x \to 5} \dfrac{10 - (x+5)}{(x-5)(x+5)} = \lim\limits_{x \to 5} \dfrac{5-x}{(x-5)(x+5)} = \\ \ \\ =\lim\limits_{x \to 5} \dfrac{-(x-5)}{(x-5)(x+5)} = \lim\limits_{x \to 5} \dfrac{-1}{x+5} = - \dfrac{1}{10}$$Another method for solving such indeterminate forms is multiplication and division by a conjugate expression. Consider the following example:

$$\lim\limits_{x \to \infty} ( \sqrt{x^2 + 2} - x)$$We multiply and divide the function $\sqrt{x^2 + 2} - x$ into the so-called conjugate expression: $\sqrt{x^2 + 2} + x$, to use the formula of the difference of squares:

$$\lim\limits_{x \to \infty} ( \sqrt{x^2 + 2} - x) = \lim\limits_{x \to \infty} \dfrac{( \sqrt{x^2 + 2} - x) \cdot ( \sqrt{x^2 + 2} + x) }{ \sqrt{x^2 + 2} + x } = \\ \ \\ = \lim\limits_{x \to \infty} \dfrac{x^2 + 2 - x^2}{\sqrt{x^2 +2} + x } = \lim\limits_{x \to \infty} \dfrac{2}{\sqrt{x^2 +2} + x } = 0$$

Conclusion

In this topic, we have learned about some indeterminate forms that arise during finding limits. We considered the $\dfrac{\infty}{\infty}, \ \dfrac{0}{0}, \ \infty - \infty$ indeterminate forms and the method with multiplication and division by a conjugate expression.