Find $$ \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{2\sqrt{2} \,\,-\, (\cos x + \sin x)^3}{1 \,\,-\, \sin 2x} $$

$$ \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{2\sqrt{2} \,\,-\, (\cos x + \sin x)^3}{1 \,\,-\, \sin 2x} $$

In numerator, $\cos x + \sin x$ is in cube form and the irrational number $2\sqrt{2}$ can also be expressed in cube form.

$$ \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{\sqrt{4 \times 2} \,\,-\, (\cos x + \sin x)^3}{1 \,\,-\, \sin 2x} $$

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{\sqrt{8} \,\,-\, (\cos x + \sin x)^3}{1 \,\,-\, \sin 2x} $$

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{\sqrt{2^3} \,\,-\, (\cos x + \sin x)^3}{1 \,\,-\, \sin 2x} $$

$\cos x + \sin x$ has cube and the number $2$ also has cube but it also has a square root additionally. Therefore, $\cos x + \sin x$ should also has square root to maintain same powers for the both expressions.

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{{(2)^\tfrac{3}{2}} \,\,-\, (\cos x + \sin x)^3}{1 \,\,-\, \sin 2x} $$

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{{(2)^\tfrac{3}{2}} \,\,-\, ((\cos x + \sin x)^{2\times\tfrac{1}{2}})^3}{1 \,\,-\, \sin 2x} $$

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{{(2)^\tfrac{3}{2}} \,\,-\, ((\cos x + \sin x)^2)^\tfrac{3}{2}}{1 \,\,-\, \sin 2x} $$

Expand $(\cos x + \sin x)^2$ to simplify it further. It can be done by using $(a + b)^2 = a^2 + b^2 + 2ab$

$\therefore \,\, (\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2 \sin x \cos x$

According to trigonometric identities

$\cos^2 x + \sin^2 x = 1$

$\sin 2x = 2 \sin x \cos x$

$\therefore \,\, (\cos x + \sin x)^2 = 1 + \sin 2x$

Now replace $(\cos x + \sin x)^2$ by $1 + \sin 2x$ in numerator of the limits expression.

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{{(2)^\tfrac{3}{2}} \,\,-\, (1 + \sin 2x)^\tfrac{3}{2}}{1 \,\,-\, \sin 2x} $$

It can be written as follows.

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{(1 + \sin 2x)^\tfrac{3}{2} \,\,-\, {(2)^\tfrac{3}{2}}}{\sin 2x \,\,-\, 1} $$

In numerator, $1+ \sin 2x$ and $2$ are two parts of the expression. Try to transform the denominator in this form. It can be done by adding $1$ and subtracting $1$ in denominator of the limits function.

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{(1 + \sin 2x)^\tfrac{3}{2} \,\,-\, {(2)^\tfrac{3}{2}}}{\sin 2x \,-\, 1 \,+\, 1 \,-\, 1} $$

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{(1 + \sin 2x)^\tfrac{3}{2} \,\,-\, {(2)^\tfrac{3}{2}}}{1 \,+\,\sin 2x \,-\, 2} $$

$$ \implies \lim_{\displaystyle x \to \dfrac{\pi}{4}} \frac{(1 + \sin 2x)^\tfrac{3}{2} \,\,-\, {(2)^\tfrac{3}{2}}}{(1 \,+\,\sin 2x) \,-\, 2} $$

The limits expression is almost transformed into following limits formula form.

Formula

$$ \lim_{\displaystyle x \to a} \frac{x^n \,-\, a^n}{x \,-\, a} \,=\, n.a^{n\,-\,1} $$

Transformation of Limits

To use this formula for our expression, limits should also form transformed.

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

$$\implies 2x \to 2 \times \frac{\pi}{4}$$

$$\implies 2x \to \frac{\pi}{2}$$

$$\implies \sin 2x \to \sin(\frac{\pi}{2})$$

$$\implies \sin 2x \to 1$$

$$\implies 1 + \sin 2x \to 1 + 1$$

$$\implies (1 + \sin 2x) \to 2$$

When $x$ tends to $\frac{\pi}{4}$, then $1 + \sin 2x$ approaches to $2$. So, change the limits of the function to simply the limits problem further.

$$ \implies \lim_{(\displaystyle 1 + \sin 2x) \to 2} \frac{(1 + \sin 2x)^\tfrac{3}{2} \,\,-\, {(2)^\tfrac{3}{2}}}{(1 \,+\,\sin 2x) \,-\, 2} $$

Assume $y = 1 + \sin 2x$ and transform the whole limits expression in terms of $y$.

$$ \therefore \,\, \lim_{\displaystyle y \to 2} \frac{y^\tfrac{3}{2} \,\,-\, {2^\tfrac{3}{2}}}{y \,-\, 2} $$

Now use the formula

$$ \implies \lim_{\displaystyle y \to 2} \frac{y^\tfrac{3}{2} \,\,-\, {2^\tfrac{3}{2}}}{y \,-\, 2} = \frac{3}{2}.(2)^{\tfrac{3}{2}\,-\, 1} $$

$$ \implies \lim_{\displaystyle y \to 2} \frac{y^\tfrac{3}{2} \,\,-\, {2^\tfrac{3}{2}}}{y \,-\, 2} = \frac{3}{2}.(2)^{\tfrac{1}{2}} $$

$$ \implies \lim_{\displaystyle y \to 2} \frac{y^\tfrac{3}{2} \,\,-\, {2^\tfrac{3}{2}}}{y \,-\, 2} = \frac{3}{2}.\sqrt{2} $$

$$ \implies \lim_{\displaystyle y \to 2} \frac{y^\tfrac{3}{2} \,\,-\, {2^\tfrac{3}{2}}}{y \,-\, 2} = \frac{3}{\sqrt2}$$

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