Question

Evaluate the following limits. $$\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{\sin ^{2}(\pi x)}$$

Short answer

Question: Evaluate the limit of the function as x approaches 2: $$\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{\sin ^{2}(\pi x)}$$ Answer: The limit of the given function as x approaches 2 is: $$\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{\sin ^{2}(\pi x)} = \frac{1}{2\pi^2}$$

Step-by-step solution

Identify the function

We are given the function: $$f(x)=\frac{x^2-4x+4}{\sin^2(\pi x)}$$ We want to evaluate the limit of this function as \(x\) approaches 2: $$\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{\sin ^{2}(\pi x)}$$

Check for direct substitution

First, try to plug in \(x=2\) directly into the function: $$f(2) = \frac{2^2 - 4\cdot 2 + 4}{\sin^2(\pi \cdot 2)} = \frac{0}{\sin^2(2\pi)} = \frac{0}{0}$$ Since we get the indeterminate form 0/0, we can't use direct substitution. We will need to apply L'Hopital's rule.

Apply L'Hopital's rule

Since we got the indeterminate form \(\frac{0}{0}\), we can use L'Hopital's rule which says: If \(\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\) is in the form \(\frac{0}{0}\), then $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$$ provided the right-hand side limit exists. For our given function, \(f(x)\) is the numerator and \(g(x)\) is the denominator: $$f(x)=x^2-4x+4$$ $$g(x)=\sin^2(\pi x)$$ Now find the derivatives of both \(f(x)\) and \(g(x)\): $$f'(x)=2x-4$$ $$g'(x)=\sin(2\pi x)\cdot 2\pi$$ Now let's plug these derivatives into our limit: $$\lim_{x\rightarrow 2} \frac{2x-4}{\sin(2\pi x)\cdot 2\pi}$$

Try direct substitution again

Now, let's try direct substitution with the new limit: $$\lim_{x\rightarrow 2} \frac{2(2)-4}{\sin(2\pi (2))\cdot 2\pi} = \frac{0}{\sin(4\pi)\cdot 2\pi} = \frac{0}{0}$$ We again get an indeterminate form. Therefore, we need to apply L'Hopital's rule one more time.

Apply L'Hopital's rule again

Since we still have the form \(\frac{0}{0}\), let's apply L'Hopital's rule again: Find the second derivatives of both \(f(x)\) and \(g(x)\): $$f''(x)=2$$ $$g''(x)=\cos(2\pi x)\cdot 4\pi^2$$ New limit with second derivatives: $$\lim_{x\rightarrow 2} \frac{2}{\cos(2\pi x)\cdot 4\pi^2}$$

Direct substitution with second derivatives

We will now use direct substitution with the second derivatives: $$\lim_{x\rightarrow 2} \frac{2}{\cos(2\pi (2))\cdot 4\pi^2} = \frac{2}{\cos(4\pi)\cdot 4\pi^2} = \frac{2}{1\cdot 4\pi^2} = \frac{1}{2\pi^2}$$ Therefore, the limit of the given function is: $$\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{\sin ^{2}(\pi x)} = \frac{1}{2\pi^2}$$

Key concepts

L'Hopital's Rule

When faced with limit problems in calculus, especially those resulting in indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hopital's Rule becomes a handy tool. This rule helps us evaluate limits by differentiating the function's numerator and denominator separately and then taking their limits.The essential condition for using L'Hopital's Rule is that both the numerator and the denominator must be differentiable at the point of interest, and the limit formed by the derivatives must exist. Here's how the rule works:

• We start with a limit \(\lim_{x\rightarrow a}\frac{f(x)}{g(x)}\) that equals \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• The rule states that \(\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}\) if the latter limit exists.

By applying derivatives, we simplify the process of finding limits, which can also involve multiple applications of the rule as seen in tougher problems.

Indeterminate Forms

Indeterminate forms occur in calculus when computations lead to ambiguous results, such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and several others. These forms indicate that further analysis is needed to determine the true behavior of limit expressions.Understanding indeterminate forms is crucial because they indicate the need for techniques beyond basic substitution. To solve these expressions, we typically:

• Attempt direct substitution first.
• If that yields an indeterminate form, explore algebraic simplification, L'Hopital's Rule, or series expansion.

Recognizing indeterminate forms is the first step in navigating through complex limit problems, guiding us to employ more sophisticated mathematical tools.

Direct Substitution

Direct substitution is the simplest method for evaluating limits in calculus. It involves substituting the value that \(x\) approaches directly into the function. If this results in a defined expression, then that is the limit.However, not all limits can be solved using this method. Often, especially with polynomial or trigonometric functions, substitution returns an indeterminate form like \(\frac{0}{0}\). This signals that we need to apply other techniques, such as L'Hopital's Rule or factorization, to resolve the limit.Despite its simplicity, direct substitution is often the first step attempted because it can quickly resolve many straightforward limit calculations without extra work.