Understanding the Limit Problem

The question asks us to find the value of the limit \[\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4x-2\pi}{\cos x}\].

First, let's evaluate the numerator and the denominator as \(x\) approaches \(\frac{\pi}{2}\):

• Numerator: As \(x \rightarrow \frac{\pi}{2}\), \(4x - 2\pi \rightarrow 4(\frac{\pi}{2}) - 2\pi = 2\pi - 2\pi = 0\).
• Denominator: As \(x \rightarrow \frac{\pi}{2}\), \(\cos x \rightarrow \cos(\frac{\pi}{2}) = 0\).

Since we have the indeterminate form \(\frac{0}{0}\), we can use methods like L'Hôpital's Rule or trigonometric substitutions to evaluate the limit.

Solving using L'Hôpital's Rule

L'Hôpital's Rule states that if \(\displaystyle\lim_{x\rightarrow c} \frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\displaystyle\lim_{x\rightarrow c} \frac{f(x)}{g(x)} = \lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}\), provided the latter limit exists.

In this case, let \(f(x) = 4x - 2\pi\) and \(g(x) = \cos x\).

• The derivative of the numerator, \(f'(x)\), is the derivative of \(4x - 2\pi\) with respect to \(x\). Since the derivative of \(4x\) is 4 and the derivative of a constant \(-2\pi\) is 0, \(f'(x) = 4\).
• The derivative of the denominator, \(g'(x)\), is the derivative of \(\cos x\) with respect to \(x\), which is \(-\sin x\).

Now, we apply L'Hôpital's Rule:

\[\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4x-2\pi}{\cos x} = \lim_{x\rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx}(4x-2\pi)}{\frac{d}{dx}(\cos x)}\]

\[= \lim_{x\rightarrow \frac{\pi}{2}} \frac{4}{-\sin x}\]

Now, we substitute \(x = \frac{\pi}{2}\) into the new expression:

\[= \frac{4}{-\sin(\frac{\pi}{2})}\]

We know that \(\sin(\frac{\pi}{2}) = 1\).

\[= \frac{4}{-1}\]

\[= -4\]

Solving using Substitution

Alternatively, we can use a substitution. Let \(y = x - \frac{\pi}{2}\). As \(x \rightarrow \frac{\pi}{2}\), \(y \rightarrow 0\). This means \(x = y + \frac{\pi}{2}\).

Substitute this into the expression:

• Numerator: \(4x - 2\pi = 4(y + \frac{\pi}{2}) - 2\pi = 4y + 2\pi - 2\pi = 4y\).
• Denominator: \(\cos x = \cos(y + \frac{\pi}{2})\). Using the trigonometric identity \(\cos(A+B) = \cos A \cos B - \sin A \sin B\), we get \(\cos(y + \frac{\pi}{2}) = \cos y \cos \frac{\pi}{2} - \sin y \sin \frac{\pi}{2} = \cos y \cdot 0 - \sin y \cdot 1 = -\sin y\).

So the limit becomes:

\[\displaystyle\lim_{y\rightarrow 0} \frac{4y}{-\sin y}\]

\[= \lim_{y\rightarrow 0} -4 \frac{y}{\sin y}\]

We know the standard limit \(\displaystyle\lim_{y\rightarrow 0} \frac{\sin y}{y} = 1\), and therefore \(\displaystyle\lim_{y\rightarrow 0} \frac{y}{\sin y} = 1\).

\[= -4 \times 1\]

\[= -4\]

Conclusion

Both methods show that the limit \[\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4x-2\pi}{\cos x}\] is equal to -4.