Limits of a function: Indeterminate Forms – Calculus

by | Sep 27, 2021 | Math Learning

While studying calculus or other branches of mathematics we may need to find the limits of a function, a sequence, or an expression, and in doing so we stumble on a situation where we cannot determine the limits, in this article we will learn about the different indeterminate forms and how to work around them in order to find the limits we are looking for.

Table of contents

  • Indeterminate Forms
  • L’Hôpital’s rule
  • Limits of a composite function
  • Limits with comparisons
  • Conclusion

 

Indeterminate Forms

We call an indeterminate form, when computing limits the case when we get an expression that we cannot determine the limit. In total there is seven indeterminate forms, here they are:

    \[ \dfrac{0}{0}; \;\; \dfrac{\infty}{\infty}; \;\; 0\times\infty; \;\; \infty-\infty; \;\; 0^{0}; \;\; 1^{\infty}; \;\; \text{ and } \infty^{0}    \]

Here are some examples to illustrate each of these indeterminate cases:

Indeterminate form \dfrac{0}{0}

    \[\lim_{x\rightarrow0}\dfrac{5-\sqrt{x^{2}+25}}{x^{2}}=\dfrac{0}{0}=\textit{ Indeterminate from }\]

Indeterminate form \dfrac{\infty}{\infty}

    \[\lim_{x\rightarrow\infty}\dfrac{\ln(x)}{x}=\dfrac{\infty}{\infty}=\textit{ Indeterminate from }\]

 

Indeterminate form 0\times\infty

    \[\lim_{x\rightarrow\infty}\sin\left(\dfrac{1}{x}\right)\times x^{2}=0\times\infty=\textit{ Indeterminate from }\]

Indeterminate form \infty-\infty

    \[\lim_{x\rightarrow0}\dfrac{1}{1-\cos(x)}-\dfrac{1}{x}=\infty-\infty=\textit{ Indeterminate from }\]

 

Indeterminate form 0^{0}

    \[\lim_{x\rightarrow0^{+}}x^{x}=0^{0}=\textit{ Indeterminate from }\]

 

Indeterminate form 1^{\infty}

    \[\lim_{x\rightarrow1^{+}}x^{\frac{1}{x-1}}=1^{\infty}=\textit{ Indeterminate from }\]

 

Indeterminate form \infty^{0}

    \[\lim_{x\rightarrow\infty}\left(1+x^{2}\right)^{\frac{1}{x}}=\infty^{0}=\textit{ Indeterminate from }\]

 

L’Hôpital’s rule and how to solve indeterminate forms

L’Hôpital’s rule is a method used to evaluate limits when we have the case of a quotient of two functions giving us the indeterminate form of the type \dfrac{0}{0}  or \dfrac{\infty}{\infty}.

The L’Hôpital rule states the following:

Theorem: L’Hôpital’s Rule:

To determine the limit of

    \[\lim_{x\rightarrow a}\dfrac{f(x)}{g(x)}\]

where a is a real number or infinity, and if we have one of the following cases:

    \[\lim_{x\rightarrow a}\dfrac{f(x)}{g(x)}=\dfrac{0}{0} \;\; \text{ or } \;\; \lim_{x\rightarrow a}\dfrac{f(x)}{g(x)}=\dfrac{\pm\infty}{\pm\infty}\]

Then we calculate the limit of the derivatives of the quotient of f and g, i.e.,

    \[\lim_{x\rightarrow a}\dfrac{f(x)}{g(x)}=\lim_{x\rightarrow a}\dfrac{f^{'}(x)}{g^{'}(x)}\]

 

Examples:

Case of \dfrac{0}{0}:

    \[\lim_{x\rightarrow 0}\dfrac{\sin(x)}{x}=\lim_{x\rightarrow 0}\dfrac{\cos(x)}{1}=\dfrac{1}{1}=1\]

 

Case of \dfrac{\pm\infty}{\pm\infty}:

In this case, after we get the derivatives of the quotient, we still get the indeterminate form of the type \dfrac{\pm\infty}{\pm\infty} so we apply L’Hôpital’s Rule again, and therefore we get:

    \[\lim_{x\rightarrow \infty}\dfrac{e^{x}}{x^{2}}=\lim_{x\rightarrow \infty}\dfrac{e^{x}}{2x}=\lim_{x\rightarrow \infty}\dfrac{e^{x}}{2}=\infty\]

 

For other Indeterminate forms, we have to do some transformation on the expression to bring it to one of the two forms that L’Hôpital’s rule solves. Let’s see some examples of how to do that!!!

L’Hôpital’s rule with the form 0\times\infty:

Let’s compute

    \[\lim_{x\rightarrow 0^{+}}x\ln(x)\]

Here we have the indeterminate form 0\times\infty, to use L’Hôpital’s rule we re-write the expression as follow:

    \[x\ln(x)=\dfrac{\ln(x)}{\dfrac{1}{x}}=\dfrac{\ln(x)}{x^{-1}}\]

Now by computing the limit we have the form \dfrac{\infty}{\infty}, therefore we can apply L’Hôpital’s rule and we get:

    \[\lim_{x\rightarrow 0^{+}}\dfrac{\ln(x)}{x^{-1}}=\lim_{x\rightarrow 0^{+}}\dfrac{\dfrac{1}{x}}{-x^{-2}}=\lim_{x\rightarrow 0^{+}}\dfrac{1}{x}(-x^{2})=\lim_{x\rightarrow 0^{+}}-x=0\]

 

L’Hôpital’s rule with the form \infty^{0}:

Let’s compute

    \[\lim_{x\rightarrow \infty}x^{\frac{1}{x}}\]

This limit gives us the form \infty^{0}, to apply the L’Hôpital’s rule we need to take a few steps as follow:

Let’s y be: y=x^{\frac{1}{x}}

By applying the natural logarithm, we get:

    \[\ln(y)=\ln\left(x^{\frac{1}{x}}\right)=\dfrac{1}{x}\ln(x)=\dfrac{\ln(x)}{x}\]

And now we compute the limit:

    \[\lim_{x\rightarrow \infty}\ln(y)=\lim_{x\rightarrow \infty}\dfrac{\ln(x)}{x}=\lim_{x\rightarrow \infty}\dfrac{\frac{1}{x}}{1}=0\]

And since we know that:

    \[e^{\ln(y)}=y\]

Therefore, we can write the limit as:

    \[\lim_{x\rightarrow \infty}x^{\frac{1}{x}}=\lim_{x\rightarrow \infty}y=\lim_{x\rightarrow \infty}e^{\ln(y)}\]

And from what we got before; we can solve the problem as follow:

    \[\lim_{x\rightarrow \infty}x^{\frac{1}{x}}=\lim_{x\rightarrow \infty}y=\lim_{x\rightarrow \infty}e^{\ln(y)}=e^{\lim_{x\rightarrow \infty}\ln(y)}=e^{0}=1\]

 

L’Hôpital’s rule with the form \infty-\infty:

Let’s compute

    \[\lim_{x\rightarrow 0^{+}}\left(\dfrac{1}{\sin x}-\dfrac{1}{x}\right)=\infty-\infty\]

This limit gives us the form \infty-\infty, to apply the L’Hôpital’s rule we need to re-write the expression, in this case, all we need to do is combine the two fractions as follow:

    \[\lim_{x\rightarrow 0^{+}}\left(\dfrac{x-\sin x}{x\sin x}\right)\]

Now the limit of the expression gives us the form \dfrac{0}{0}. Now by applying the L’Hôpital’s rule twice (because we get the indeterminate form \dfrac{0}{0} after the first time) we get:

    \[\lim_{x\rightarrow 0^{+}}\left(\dfrac{x-\sin x}{x\sin x}\right)=\lim_{x\rightarrow 0^{+}}\left(\dfrac{1-\cos x}{\sin x + x\cos x}\right)=\lim_{x\rightarrow 0^{+}}\left(\dfrac{\sin x}{2\cos x - x\sin x}\right)=\dfrac{0}{2-0}=0\]

 

L’Hôpital’s rule with the form 1^{\infty}:

Let’s compute

    \[\lim_{x\rightarrow \infty}\left(1+\dfrac{4}{x}\right)^{x}\]

This limit gives us the form 1^{\infty}, to avoid it and be able to apply L’Hôpital’s rule we need to re-write the expression as follow:

Let y=\left(1+\dfrac{4}{x}\right)^{x}

Then

    \[\lim_{x\rightarrow \infty}\ln(y)=\lim_{x\rightarrow \infty}\ln\left(1+\dfrac{4}{x}\right)^{x}=\lim_{x\rightarrow \infty}x\ln\left(1+\dfrac{4}{x}\right)=\lim_{x\rightarrow \infty}\dfrac{\ln\left(1+\dfrac{4}{x}\right)}{\dfrac{1}{x}}\]

Using L’Hôpital’s rule we get:

    \[\lim_{x\rightarrow \infty}\dfrac{\dfrac{1}{1+\frac{4}{x}}\left(\dfrac{-4}{x^{2}}\right)}{-\dfrac{1}{x^{2}}}=\lim_{x\rightarrow \infty}\dfrac{4}{1+\dfrac{4}{x}}=4\]

And therefore, we get:

    \[\lim_{x\rightarrow \infty}\left(1+\dfrac{4}{x}\right)^{x}=e^{4}\]

 

L’Hôpital’s rule with the form 0^{0}:

Let’s compute

    \[\lim_{x\rightarrow 0^{+}}x^{x}\]

This limit gives us the indeterminate form 0^{0}, to use the L’Hôpital’s rule we need to re-write the expression as follow:

    \[x^{x}=e^{x\ln(x)}\]

Now we calculate the limit of the exponent using L’Hôpital’s rule:

    \[\lim_{x\rightarrow 0^{+}}x\ln(x)=\lim_{x\rightarrow 0^{+}}\dfrac{\ln(x)}{\dfrac{1}{x}}=\lim_{x\rightarrow 0^{+}}\dfrac{\frac{1}{x}}{-\frac{1}{x^{2}}}=\lim_{x\rightarrow 0^{+}}-x=0\]

Therefore,

    \[\lim_{x\rightarrow 0^{+}}x^{x}=\lim_{x\rightarrow 0^{+}}e^{x\ln(x)}=e^{0}=1\]

 

Limits of a composite function

Theorem:

Let a, b and c represent real numbers or +\infty or -\infty, and let u, v, and f be functions that verify f=v\circ u.

If the limit of the function u when x tends to a is b, and the limit of the function v when x tend to b is c then the limit of the function f when x tends to a is c.

Meaning: if \displaystyle \lim_{x\rightarrow a}u(x)=b and if \displaystyle \lim_{x\rightarrow b}v(x)=c then \displaystyle \lim_{x\rightarrow a}f(x)=c

Example:

Let’s consider the function f defined on the domain \mathbb{R^{\ast}} as

    \[f(x)=\sin\left(\dfrac{\pi}{2}+\dfrac{1}{x}\right)\]

and we want to determine the limit of the function f when x tends to +\infty, i.e., \displaystyle \lim_{x\rightarrow +\infty}f(x)

We notice that the function f is a composite of two functions, precisely f is a composite of the functions u and v in this order (f=v\circ u), where

    \[u(x)=\dfrac{\pi}{2}+\dfrac{1}{x}\]

and

    \[v(x)=\sin x\]

Since

    \[\lim_{x\rightarrow +\infty} u(x)=\dfrac{\pi}{2}\]

And

    \[\lim_{x\rightarrow \dfrac{\pi}{2}} v(x)=1\]

Therefore

    \[\lim_{x\rightarrow +\infty}f(x)=1\]

 

Limits with comparisons

Theorem 1:

Suppose f, g, and h three functions, and l a real number; if we have \displaystyle \lim_{x\rightarrow +\infty}g(x)=l and \displaystyle \lim_{x\rightarrow +\infty}h(x)=l and if for x big enough we have g(x)\leq f(x)\leq h(x) then \displaystyle \lim_{x\rightarrow +\infty}f(x)=l.

 

Example:

Let’s consider the function f defined on \mathbb{R^{\ast}}  as \displaystyle \lim_{x\rightarrow +\infty}f(x)=\dfrac{\sin x}{x}

We know that for every x from \mathbb{R}, we have \displaystyle -1\leq \sin x \leq 1

And therefore, for every x in ]0;+\infty[, we have -\dfrac{1}{x}\leq \dfrac{\sin x}{x}\leq\dfrac{1}{x}

And since \displaystyle \lim_{x\rightarrow +\infty}\left(-\dfrac{1}{x}\right)=\lim_{x\rightarrow +\infty}\left(\dfrac{1}{x}\right)=0 we conclude that  \displaystyle \lim_{x\rightarrow +\infty}\left(\dfrac{\sin x}{x}\right)=0

 

 

Theorem 2:

Suppose f and g two functions and l a real number; if we have \displaystyle \lim_{x\rightarrow +\infty}g(x)=+\infty, and if for x big enough we have f(x)\geq g(x) then \displaystyle \lim_{x\rightarrow +\infty}f(x)=+\infty.

 

Theorem 3:

Suppose f and g two functions and l a real number; if we have \displaystyle \lim_{x\rightarrow +\infty}g(x)=-\infty, and if for x big enough we have f(x)\leq g(x) then \displaystyle \lim_{x\rightarrow +\infty}f(x)=-\infty.

 

Remarque: these three theorems can be extended to the two cases for the limit when x tends to -\infty or a real number.

 

Example:

Let’s consider the function f defined on \mathbb{R} as f(x)=x+\sin x

We know that for every x from \mathbb{R}, we have -1\leq \sin x \leq 1, and then for every x from \mathbb{R}, we have

    \[x-1\leq x+\sin x\leq x+1\]

Therefore:

Since

    \[\lim_{x\rightarrow +\infty}(x-1)=+\infty\]

Then

    \[\lim_{x\rightarrow +\infty}f(x)=+\infty\]

And since

    \[\lim_{x\rightarrow -\infty}(x+1)=-\infty\]

Then

    \[\lim_{x\rightarrow -\infty}f(x)=-\infty\]

 

Conclusion

In this article, we discovered the different indeterminate forms and how to avoid them and calculate the limits using L’Hôpital’s rule, with examples of the various cases. Also, we learned about how to determine the limits of composite function and how to determine limits with comparison. Don’t miss the previous articles about the idea of limits, their properties, and the arithmetic operations on them.

Also, if you want to learn more fun subjects, check the post about Functions and some of their properties, or the one about How to solve polynomial equations of first, second, and third degrees!!!!!

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