Question

$\lim_{x\to\frac{\pi}{4}}\frac{\tan x-1}{\cos 2x}$ is equal to

Answer 1

-1

Answer 2

The limit to be evaluated is $\lim_{x\to\frac{\pi}{4}}\frac{\tan x-1}{\cos 2x}$.

Substituting $x = \frac{\pi}{4}$ into the expression, we get:

Numerator: $\tan(\frac{\pi}{4}) - 1 = 1 - 1 = 0$.

Denominator: $\cos(2 \times \frac{\pi}{4}) = \cos(\frac{\pi}{2}) = 0$.

The limit is of the indeterminate form $\frac{0}{0}$. We can use L'Hopital's Rule or algebraic manipulation.

Method 1: Using L'Hopital's Rule

Let $f(x) = \tan x - 1$ and $g(x) = \cos 2x$.

Then $f'(x) = \frac{d}{dx}(\tan x - 1) = \sec^2 x$.

And $g'(x) = \frac{d}{dx}(\cos 2x) = -\sin(2x) \cdot \frac{d}{dx}(2x) = -2\sin(2x)$.

Applying L'Hopital's Rule, the limit is:

$\lim_{x\to\frac{\pi}{4}}\frac{f'(x)}{g'(x)} = \lim_{x\to\frac{\pi}{4}}\frac{\sec^2 x}{-2\sin(2x)}$.

Now, substitute $x = \frac{\pi}{4}$ into this new expression:

$\frac{\sec^2(\frac{\pi}{4})}{-2\sin(2 \times \frac{\pi}{4})} = \frac{(\frac{1}{\cos(\frac{\pi}{4})})^2}{-2\sin(\frac{\pi}{2})} = \frac{(\frac{1}{1/\sqrt{2}})^2}{-2(1)} = \frac{(\sqrt{2})^2}{-2} = \frac{2}{-2} = -1$.

Method 2: Using algebraic manipulation

We can rewrite the expression using trigonometric identities.

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

We can factor out $-1$ from the numerator term $(\sin x - \cos x)$ to get $-(\cos x - \sin x)$.

So the expression becomes:

$\frac{-(\cos x - \sin x)}{\cos x (\cos x - \sin x)(\cos x + \sin x)}$.

For $x \neq \frac{\pi}{4}$, $\cos x - \sin x \neq 0$, so we can cancel the term $(\cos x - \sin x)$ from the numerator and the denominator:

$\frac{-1}{\cos x (\cos x + \sin x)}$.

Now, evaluate the limit as $x \to \frac{\pi}{4}$:

$\lim_{x\to\frac{\pi}{4}}\frac{-1}{\cos x (\cos x + \sin x)}$.

Substitute $x = \frac{\pi}{4}$:

$\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

$\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

The denominator becomes $\cos(\frac{\pi}{4}) (\cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4})) = \frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} (\frac{2}{\sqrt{2}}) = \frac{1}{\sqrt{2}} (\sqrt{2}) = 1$.

The limit is $\frac{-1}{1} = -1$.

Both methods yield the same result.