 # L’Hôpital’s Rule Calculator + Online Solver With Free Steps

L’Hopital’s Rule Calculator is used to solve Indeterminate Form problems of limits for algebraic expressions in analytics and calculus. The Calculator can take in input details regarding the algebraic expression as well as the limit applied to it.

An Indeterminate Form of a limit for any type of algebraic expression refers to a combination of a numerator and denominator functions. This mathematical combination results in an output that does not have a lot of meaning attached to it, or is either infinite or a ratio of zeros and infinities.

This sort of output is the mathematical expression of an Indeterminate Form’s unviable solution. ## What Is the L’Hospital’s Rule Calculator?

L’Hopital’s Rule Calculator is an online tool that you can use to solve your indeterminate limit problems.

This calculator is very helpful and comes in handy as you can simply just use it in your browser.

An Indeterminate form of any algebraic expression with a limit applied to it corresponds to a solution that is not viable. Thus, it is of the form zero over zero, zero over infinity, and so on and therefore does not provide insight into the problem appropriately.

This is another one of the more Advanced Calculators we offer, as it solves algebra and calculus problems for you in real-time.

## How To Use the L’Hospital’s Rule Calculator?

To use L’Hopital’s Rule Calculator, you must first have a mathematical expression with a limit that does not provide a viable solution. Such an expression is referred to as the Indeterminate Form of the Limit.

Once you have a problem setup, place the values of this problem into input boxes provided in the calculator and solve your L’Hopital’s Rule problem. Now, to get the best results from your Indeterminate problems of Limit, you may follow the given step-by-step instructions:

### Step 1

You can start by entering the limit’s value inside the calculator. This is done at the input box labeled as “lim x to” by plugging in the limit information.

### Step 2

Following the first step, you may enter the function definition of your algebraic equation in another input box. This input box is labeled as “of” as it represents the expression to be solved for the given limit.

### Step 3

The next step includes pressing the “Submit” button, which will start solving your problem and a solution will present itself.

### Step 4

The last and final step of the use of this calculator is the reuse capabilities, you can simply solve more problems using this calculator if you wish to. This is done by setting up more problems and entering their values into the input boxes.

An important fact to note about this calculator is that it can be used for a lot of different types of expressions. It can solve for algebraic, trigonometric, exponential, and so on types of expressions. This fact only adds to its value and its capacity to handle different kinds of problems.

## How Does the L’Hôpital’s Rule Calculator Work?

L’Hopital’s Rule Calculator works by using a set of techniques expressed as the L’Hopital’s Rule to convert a seemingly indeterminate problem into a determinate solution. Thus, getting a solution for an unsolvable problem is the process known as L’Hopital’s Rule.

### Limits in Algebra

Limits are used in algebra to showcase or even study the behavior of a Function as the input to that function, as it changes and approaches a certain value. They are avidly used in Analytics and Calculus, as they come in very handy when dealing with systems that change rapidly based on input values.

The Limits help break down the trend for the pattern in the outputs of a function. Now, take a general function of the form:

$y = \lim_{x \to c} f(x)$

Considering that, at $x = c-10$ the solution to $f(x)$ starts getting very big, and it keeps getting big until $x = c-1$. But then, at $x = c$ the solution goes to infinity, the limit isn’t going to provide the best result to our problem, given as:

$y = \lim_{x \to c} f(x) = f(x) \to \infty$

### Indeterminate Form

The Indeterminate Form of limits for algebraic expressions is referred to as an expression dealing with a couple of functions which do not have a viable solution to them based on their limits.

So, consider the given generalized function as a limit problem:

$f(x) = \frac{g(x)}{h(x)}$

$y = \lim_{x \to c} \frac{g(x)}{h(x)}$

So the output of this Limit can lead to a bunch of indeterminate solutions i.e., zero over zero, zero over infinity, or vice versa, etc. In general, a total of seven different Indeterminate Outcomes are possible from this problem.

$\begin{matrix} \frac{0}{0} & \frac{\infty}{\infty} & 0 \cdot \infty & 0^0 & \infty^0 & 1^{\infty} & \infty – \infty \end{matrix}$

### Solve for Indeterminate Form

For solving the Indeterminate Form of limit, Transformations are used which helps in finding a form that is not indeterminate for the given combination of functions.

We have come across solutions involving the Common Denominator technique, Trigonometric Identities, Conjugate Pairs, Horizontal Asymptote Calculation, and Factorization for Indeterminate Form. But often, one may find themselves struggling with calculating the desired solution, even when using one of these solutions.

Therefore, we are going to use L’Hopital’s Rule for solving Indeterminate Form for limits applied to algebraic equations.

### L’Hopital’s Rule

The L’Hopital’s Rule describes that if the numerator and denominator functions of an indeterminate form are differentiable, then their derivatives’ ratio corresponds to their original ratio’s output. This fact is expressed mathematically as follows:

$\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$

This is mainly true for indeterminate form, which can be expressed as:

$\lim_{x \to c} \frac{g(x)}{h(x)} = \frac{0}{0} , \lim_{x \to c} \frac{g'(x)}{h'(x)} = \frac{\infty}{\infty}$

One thing to note is that sometimes, one may have to apply L’Hopital’s Rule more than once to get the desired results, which means acquiring a limited solution.

## History of L’Hôpital’s Rule

The solution to Indeterminate limit expressions holds a very significant place in the mathematical world. And so does the L’Hospital’s Rule, as it is used for finding solutions to complex seemingly unsolvable problems.

The L’Hôpital’s Rule is therefore one of the most significant solution methods of mathematics, or more specifically Calculus. The L’Hôpital’s Rule also referred to as L’Hospital’s Rule is more commonly known as Bernoulli’s Rule. It is used to calculate the mathematically indeterminate algebraic expressions’ limits.

The rule works by solving its problem’s indeterminate solution any number of times, to make it solvable. This means that it could be used more than once when attempting to find the determinate solution. It was named L’Hospital’s Rule in the memory of the French Mathematician Guillaume de L’Hopital from the 17th century.

Although it was named after L’Hopital as a tribute for his work in the making of this. But the baseline for this rule, Bernoulli’s Rule was brought forward to L’Hopital by Johann Bernoulli back in 1694.

## Solved Examples

Here are some examples solved using this calculator.

### Example 1

Consider the given limit equation:

$\lim_{x \to \infty} \frac{2x^2 – 5x + 1}{ 3 + x + 6x^2}$

Solve the solution to the algebraic equation for the given limit using L’Hospital’s Rule.

### Solution

We start by testing the algebraic expression and whether it produces a determinate solution.

For that, we will put in the given value of the limit and solve the expression for it.

$\frac{2(\infty)^2 – 5(\infty) + 1}{ 3 + (\infty) + 6(\infty)^2} = \frac{\infty}{\infty}$

Therefore, we conclude that this algebraic expression is in the indeterminate form for the limit approaches to infinity.

Now, we move forward by applying L’Hospital’s Rule to our algebraic expression.

The expression becomes:

$\lim_{x \to \infty} \frac{\frac{d}{dx}(2x^2 – 5x + 1)}{\frac{d}{dx} (3 + x + 6x^2)} = \lim_{x \to \infty} \frac{4x-5}{1+12x}$

Solving this resulting expression for the limit approaching to infinity results in the following equation:

$\lim_{x \to \infty} \frac{4x-5}{1+12x} = \frac{4(\infty)-5}{1+12(\infty)} = \frac{\infty}{\infty}$

This goes to show that this problem is still indeterminate, therefore we must proceed by applying the L’Hospital’s Rule again:

$\lim_{x \to \infty} \frac{\frac{d}{dx}(4x-5)}{\frac{d}{dx}(1+12x)} = \lim_{x \to \infty} \frac{4}{12}$

$\lim_{x \to \infty} \frac{4}{12} = \frac{4}{12} = \frac{1}{3}$

Thus, in this case, we applied the L’Hospital’s Rule twice to get our desired result which was a determinate form for the original limit.

### Example 2

Now consider the given limit equation:

$\lim_{x \to 0^+} \sin x ^ {\tan x}$

Solve the solution to the trigonometric equation for the given limit using L’Hospital’s Rule.

### Solution

We begin just like before, by testing the trigonometric expression. To see whether it can produce a determinate solution.

Thus, we plug in the value of the limit applied to the expression and compute the result.

$\lim_{x \to 0^+} \sin x ^ {\tan x} = (\sin (0)) ^ {(\tan (0))} = 0^{\infty}$

The test proves that the trigonometric expression is indeed indeterminant as the output was zero to the power of infinity.

Now, we must carry on by applying the L’Hospital’s Rule on this trigonometric expression.

But for doing that, we apply a transformation to this expression first to convert this into a quotient.

$y = (\sin x )^ {\tan x}$

$\ln y = \ln (\sin x )^ {\tan x}$

$\ln y = \tan x \cdot \ln (\sin x )$

$\ln y = \frac{1}{\cot x} \cdot \ln (\sin x )$

$\ln y = \frac {\ln (\sin x )}{\cot x}$

This is our resulting transformed trigonometric expression. Now we can apply the differentials on it.

$\ln y = \lim_{x \to 0^+} \frac {\ln (\sin x )}{\cot x}$

$\ln y = \lim_{x \to 0^+} \frac {\frac{d}{dx}(\ln (\sin x ))}{\frac{d}{dx}(\cot x)} = \lim_{x \to 0^+} \frac {\frac{\cos x}{\sin x}}{-\csc ^2 x}$

$\ln y = \lim_{x \to 0^+} \frac {\frac{\cos x}{\sin x}}{-\csc ^2 x} = \lim_{x \to 0^+} (-\cos x \cdot \sin x) = (-\cos(0) \cdot \sin(0)) = -1 \cdot 0 = 0$

Although this solution also looks indeterminate as the result came to be zero, but solving for the natural logarithm applied on $y$ we would get this result:

$\ln y = 0$

$e ^ {\ln y} = e ^ 0$

$y = e ^ 0 = 1$

The final result is thus a determinate solution, which we were able to extract from the indeterminate form by solving via L’Hospital’s Rule.

### Example 3

Consider another problem involving a trigonometric expression given as:

$\lim_{x \to 0^+} (\csc x – \cot x)$

Now solve the expression for the given limit.

### Solution

We shall begin by setting up the test for indeterminate form, this is done by plugging the limit value into the expression. Solving that will give us a concrete idea of whether the limit is in its indeterminate form.

$\lim_{x \to 0^+} (\csc x – \cot x) = (\csc (0) – \cot (0)) = \infty – \infty$

The result is indeed an indeterminate solution to our limit problem.

Therefore, we will move forward by transforming the initial trigonometric expression into their trigonometric identities, in general conduct simplification. This is done as follows:

$\lim_{x \to 0^+} (\csc x – \cot x) \Leftrightarrow \lim_{x \to 0^+} \bigg(\frac{1}{\sin x} – \frac{\cos x}{\sin x} \bigg) \Leftrightarrow \lim_{x \to 0^+} \bigg(\frac{1 – \cos x}{\sin x} \bigg)$

Here we have the transformed result for the trigonometric expression, and it can now be run through the L’Hospital’s Rule to get the result. The L’Hospital’s Rule is applied as follows:

$\lim_{x \to 0^+} \bigg(\frac{\frac{d}{dx}(1 – \cos x)}{\frac{d}{dx}(\sin x)} \bigg) = \lim_{x \to 0^+} \frac{-(-\sin x)}{\cos x}$

$\lim_{x \to 0^+} \frac{\sin x}{\cos x} = \frac{\sin(0)}{\cos(0)} = \frac{0}{1} = 0$

And the solution is in its determinate form, it must be noted that this is not an indeterminate form, as the zero output corresponds to a number rather than the seven types of indeterminate solutions. And this is found using the L’Hospital’s Rule as usual.