Question

$\lim_{x \to \frac{\pi}{4}}(\cos x - \sin x)^{(x-\frac{\pi}{4})}$ is equal to :

Answer 1

1

Answer 2

To evaluate the limit $\lim_{x \to \frac{\pi}{4}}(\cos x - \sin x)^{(x-\frac{\pi}{4})}$, first determine its form.

As $x \to \frac{\pi}{4}$:

The base $\cos x - \sin x \to \cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$.

The exponent $x - \frac{\pi}{4} \to \frac{\pi}{4} - \frac{\pi}{4} = 0$.

So, the limit is of the indeterminate form $0^0$.

To solve limits of the form $f(x)^{g(x)}$, we use the property $f(x)^{g(x)} = e^{g(x)\ln f(x)}$.

Let $L = \lim_{x \to \frac{\pi}{4}}(\cos x - \sin x)^{(x-\frac{\pi}{4})}$.

Then $\ln L = \lim_{x \to \frac{\pi}{4}} (x-\frac{\pi}{4}) \ln(\cos x - \sin x)$.

For $\ln(\cos x - \sin x)$ to be defined in real numbers, the argument $(\cos x - \sin x)$ must be positive.

$\cos x - \sin x > 0 \implies \cos x > \sin x$.

In the vicinity of $x = \frac{\pi}{4}$, this condition holds for $x < \frac{\pi}{4}$. Therefore, we must consider the left-hand limit, i.e., $\lim_{x \to (\frac{\pi}{4})^-}$.

Let $x = \frac{\pi}{4} - h$, where $h \to 0^+$.

Then $x - \frac{\pi}{4} = -h$.

And $\cos x - \sin x = \cos(\frac{\pi}{4} - h) - \sin(\frac{\pi}{4} - h)$.

Using the compound angle formulas:

$\cos(\frac{\pi}{4} - h) = \cos\frac{\pi}{4}\cos h + \sin\frac{\pi}{4}\sin h = \frac{1}{\sqrt{2}}(\cos h + \sin h)$.

$\sin(\frac{\pi}{4} - h) = \sin\frac{\pi}{4}\cos h - \cos\frac{\pi}{4}\sin h = \frac{1}{\sqrt{2}}(\cos h - \sin h)$.

Substituting these into the expression for the base:

$\cos x - \sin x = \frac{1}{\sqrt{2}}(\cos h + \sin h) - \frac{1}{\sqrt{2}}(\cos h - \sin h)$ $= \frac{1}{\sqrt{2}}(\cos h + \sin h - \cos h + \sin h)$ $= \frac{1}{\sqrt{2}}(2\sin h) = \sqrt{2}\sin h$.

Now, substitute these into the limit expression for $\ln L$:

$\ln L = \lim_{h \to 0^+} (-h) \ln(\sqrt{2}\sin h)$.

Using logarithm properties, $\ln(AB) = \ln A + \ln B$:

$\ln L = \lim_{h \to 0^+} (-h) [\ln\sqrt{2} + \ln(\sin h)]$.

$\ln L = \lim_{h \to 0^+} [-h\ln\sqrt{2} - h\ln(\sin h)]$.

We evaluate each term separately:

1. $\lim_{h \to 0^+} (-h\ln\sqrt{2}) = 0 \times \ln\sqrt{2} = 0$.

2. $\lim_{h \to 0^+} (-h\ln(\sin h))$. This is of the form $0 \times (-\infty)$ as $h \to 0^+$ and $\sin h \to 0^+$, so $\ln(\sin h) \to -\infty$.

To apply L'Hopital's Rule, rewrite it as a fraction:

$\lim_{h \to 0^+} \frac{-\ln(\sin h)}{1/h}$. This is of the form $\frac{\infty}{\infty}$.

Apply L'Hopital's Rule:

$\lim_{h \to 0^+} \frac{\frac{d}{dh}(-\ln(\sin h))}{\frac{d}{dh}(1/h)} = \lim_{h \to 0^+} \frac{-\frac{\cos h}{\sin h}}{-\frac{1}{h^2}}$. $= \lim_{h \to 0^+} \frac{h^2 \cos h}{\sin h}$.

This is of the form $\frac{0}{0}$. We can rewrite it as:

$\lim_{h \to 0^+} \left(\frac{h}{\sin h} \cdot h \cos h\right)$.

We know that $\lim_{h \to 0^+} \frac{\sin h}{h} = 1$, so $\lim_{h \to 0^+} \frac{h}{\sin h} = 1$.

Also, $\lim_{h \to 0^+} h \cos h = 0 \times \cos(0) = 0 \times 1 = 0$.

Therefore, $\lim_{h \to 0^+} (-h\ln(\sin h)) = 1 \times 0 = 0$.

Combining the results for both terms:

$\ln L = 0 + 0 = 0$.

Since $\ln L = 0$, we have $L = e^0 = 1$.

The final answer is $\boxed{1}$.

Explanation of the solution:

1. Identify the indeterminate form $0^0$.
2. Use logarithm: $L = \lim_{x \to a} f(x)^{g(x)} \implies \ln L = \lim_{x \to a} g(x) \ln f(x)$.
3. Analyze the domain of $\ln(\cos x - \sin x)$, which requires $\cos x - \sin x > 0$. This implies taking the left-hand limit as $x \to \frac{\pi}{4}$.
4. Substitute $x = \frac{\pi}{4} - h$ and simplify the expression.
5. Evaluate $\lim_{h \to 0^+} (-h) [\ln\sqrt{2} + \ln(\sin h)]$ term by term.
6. The first term $-h\ln\sqrt{2} \to 0$.
7. The second term $-h\ln(\sin h)$ is of $0 \times (-\infty)$ form, convert to $\frac{\infty}{\infty}$ form and apply L'Hopital's rule, or use standard limits like $\lim_{h \to 0} \frac{\sin h}{h} = 1$. This term also evaluates to 0.
8. So, $\ln L = 0$, which means $L = e^0 = 1$.