lim(x→π/4) (8√2 - (cos x + sin x)⁷) / (√2 - √2 sin 2x) is equal to

Q: Why did we choose $t = \sin x + \cos x$ as our substitution?

This is a classic JEE strategy! When you see $\sin x + \cos x$ and $\sin 2x$ together, it is a huge hint. Squaring $t = \sin x + \cos x$ gives $t^2 = 1 + \sin 2x$, which perfectly links the two parts of the expression. It turns a complex trigonometric limit into a simple polynomial one.

Q: Why does the limit boundary change from $x \to \frac{\pi}{4}$ to $t \to \sqrt{2}$?

When you change the variable of a limit, you must also change the value that the variable approaches. Since $t = \sin x + \cos x$, as $x$ approaches $\frac{\pi}{4}$, $t$ approaches $\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Always remember to update your limit boundary!

Q: Can we use series expansion instead of L'Hopital's Rule?

Yes, you absolutely can! You could expand the terms around $x = \frac{\pi}{4}$ using Taylor series. However, for this specific problem, the algebraic substitution followed by L'Hopital's Rule is much faster and less prone to calculation errors. Both methods are valid, but choose the one that feels most intuitive to you.

Question

$\lim_{x 	o \frac{\pi}{4}} \frac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2} \sin 2x}$ is equal to

Options:

(A) 14

(B) 7

(C) 14$\sqrt{2}$

(D) 7$\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given limit is:
$$\lim_{x 	o \frac{\pi}{4}} \frac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2} \sin 2x}$$

At first glance, this expression appears intimidating. However, in the context of JEE Advanced, complex problems are often simple ones in disguise. The key is to identify the underlying structure.

### The First Step: Recognizing the Pattern

When you observe $(\sin x + \cos x)$ paired with $\sin 2x$, your mathematical intuition should immediately identify the identity:
$$(\sin x + \cos x)^2 = 1 + \sin 2x$$

This is the secret key to unlocking the problem. By substituting $t = \sin x + \cos x$, we translate a difficult trigonometric expression into the much more manageable language of algebra.

### The Transformation

As $x 	o \frac{\pi}{4}$, our new variable $t$ approaches $\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \sqrt{2}$.

Now, consider the denominator. Since $\sin 2x = t^2 - 1$, we substitute this into the limit:
$$\lim_{t 	o \sqrt{2}} \frac{8\sqrt{2} - t^7}{\sqrt{2} - \sqrt{2}(t^2 - 1)}$$

### Simplifying the Landscape

Let us clean up the denominator. Factoring out $\sqrt{2}$, we obtain $\sqrt{2}[1 - (t^2 - 1)]$, which simplifies to $\sqrt{2}(2 - t^2)$.

The limit now becomes:
$$\lim_{t 	o \sqrt{2}} \frac{8\sqrt{2} - t^7}{\sqrt{2}(2 - t^2)}$$

If you evaluate this at $t = \sqrt{2}$, you obtain the $\frac{0}{0}$ indeterminate form. We have successfully reduced a trigonometric nightmare to a simple polynomial ratio.

### The Final Blow: L'Hopital's Rule

Since we have a $\frac{0}{0}$ form, we apply **L'Hopital's Rule** by differentiating the numerator and the denominator with respect to $t$.

The derivative of the numerator $8\sqrt{2} - t^7$ is $-7t^6$. The derivative of the denominator $\sqrt{2}(2 - t^2)$ is $-2\sqrt{2}t$.

The limit is now:
$$\lim_{t 	o \sqrt{2}} \frac{-7t^6}{-2\sqrt{2}t} = \lim_{t 	o \sqrt{2}} \frac{7t^5}{2\sqrt{2}}$$

### The Victory

Finally, we substitute $t = \sqrt{2}$ into our simplified expression:
$$\frac{7(\sqrt{2})^5}{2\sqrt{2}}$$

Since $(\sqrt{2})^5 = 4\sqrt{2}$, the expression becomes:
$$\frac{7 \cdot 4\sqrt{2}}{2\sqrt{2}} = \frac{28}{2} = 14$$

The monster has been tamed. The final answer is **14**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to \frac{π}{4}} \frac{8 2 - ( c o s x + s i n x ) ^{7}}{2 - 2 s i n 2 x}$ is equal to

Select Answer:

Visualized Solution

Analyze the Limit Expression

Substitution $t = sin x + cos x$

Finding the New Limit for $t$

Relating $sin 2 x$ to $t$

Substitute $t$ into the Limit

Simplify the Denominator

Check for Indeterminate Form

Apply L'Hopital's Rule

Differentiate the Numerator

Differentiate the Denominator

Simplify the Derivative Fraction

Substitute $t = 2$

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The First Step

The Transformation

Simplifying the Landscape

The Final Blow

The Victory

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→4π​​2​−2​sin2x82​−(cosx+sinx)7​ is equal to