L'Hopital's Rule: Solving Indeterminate Limits
Introduction
In the world of calculus, there are instances where mathematical problems seem to defy straightforward solutions. One such conundrum arises when you encounter a limit of a quotient where both the numerator and denominator approach zero as a variable, in this case, h, gets infinitely small. To tackle this issue, mathematicians have devised a powerful tool known as L’Hopital’s Rule. In this blog, we will dive into the mechanics of this rule, but first, let’s examine the problem that necessitates its application.
The Limit
Consider the following limit:
\[\lim_{h \to 0}\frac{\tan(2(x + h)) - \tan(2x)}{h}\]This expression is intriguing because, as h approaches zero, both the numerator and the denominator also approach zero. When we encounter such an indeterminate form (0/0), it leaves us scratching our heads. How can we make sense of this seemingly impossible situation?
Understanding the Numerator
Let’s delve into the numerator to gain a better understanding of this limit. The numerator can be expressed as:
\(\tan(2(x + h)) - \tan(2x)\)
This expression involves the subtraction of two tangent functions. One of these functions pertains to the quantity \(2(x + h)\), while the other corresponds to \(2x\). When we scrutinize the behavior of this difference as \(h\) approaches zero, we observe a fascinating phenomenon.
As \(h\) diminishes to infinitesimal levels, the angles inside the tangent functions draw nearer to each other. This convergence leads to a difference that can be precisely calculated. In essence, as \(h\) tends towards zero, the limit of this expression indeed evaluates to zero. Consequently, the numerator converges to zero as \(h\) approaches zero.
Understanding the Denominator
Next, let’s analyze the behavior of the denominator:
\[h\]The denominator, on the other hand, is quite straightforward. It’s simply h, and as h approaches zero, the denominator indeed becomes zero.
Introducing L’Hopital’s Rule
Now, we find ourselves in a situation where the numerator doesn’t vanish as h approaches zero, while the denominator does. This is where L’Hopital’s Rule comes to our rescue. Named after the French mathematician Guillaume de l’Hopital, this rule offers a method for evaluating indeterminate limits of the form 0/0 and \(\infty/\infty\).
L’Hopital’s Rule states that if the limit of a ratio (0/0 or \(\infty/\infty\)) exists, it is equal to the limit of the ratio of the derivatives of the numerator and denominator:
\[\lim_{h \to 0}\frac{f(x + h) - f(x)}{g(x + h) - g(x)} = \lim_{h \to 0}\frac{f'(x)}{g'(x)}\]Calculating the Derivatives
Now, let’s calculate the derivatives of the numerator and denominator.
For the numerator, which is \(\tan(2(x + h)) - \tan(2x)\), we need to find the derivative with respect to h. The derivative of \(\tan(2(x + h))\) is \(\sec^2(2(x + h)) \cdot 2\), and the derivative of \(\tan(2x)\) is \(\sec^2(2x) \cdot 0\) because it’s a constant with respect to h. So, the derivative of the numerator is:
\[2\sec^2(2(x + h))\]For the denominator, which is simply \(h\), its derivative with respect to h is:
\[1\]Now, we can apply L’Hopital’s Rule to our original limit:
\[\lim_{h \to 0}\frac{\tan(2(x + h)) - \tan(2x)}{h} = \lim_{h \to 0}\frac{2\sec^2(2(x + h))}{1}\]As h approaches zero, the limit becomes:
\[2\sec^2(2x)\]Conclusion
By applying L’Hopital’s Rule and correctly calculating the derivatives of the numerator and denominator, we’ve successfully evaluated the initially perplexing limit. It simplifies to \(2\sec^2(2x)\), giving us a clear and non-indeterminate result. L’Hopital’s Rule is a valuable tool in solving limits with indeterminate forms, and mastering it can unlock the solutions to complex problems in calculus.