Analyzing the Setup
Imagine you are standing before a complex limit expression, staring at a jumble of trigonometric and logarithmic functions. It looks like a chaotic mess, but in the world of JEE Advanced, chaos is just an opportunity for order.
We are evaluating the following limit:
x→0lim((cos3(4x)1−cos2(3x))((loge(2x+1))5sin3(4x)))
Phase 1
The First Simplification
The first thing that should catch your eye is the term 1−cos2(3x). It is a classic trigonometric identity in disguise.
We know that sin2(θ)+cos2(θ)=1, which implies 1−cos2(θ)=sin2(θ). By setting θ=3x, our numerator instantly transforms into sin2(3x).
The expression now looks much friendlier:
x→0limcos3(4x)sin2(3x)⋅(loge(1+2x))5sin3(4x)
Phase 2
The Standard Limit Toolkit
Now, we need our secret weapons. We rely on two fundamental limits:
u→0limusinu=1andu→0limuloge(1+u)=1
Notice that cos3(4x) is harmless. As x→0, cos(4x)→1, so cos3(4x)→1. It will not cause any division-by-zero issues.
Phase 3
The Balancing Act
This is the most crucial part of the journey. To use our standard limits, we must force the denominators to match the arguments of our functions.
For sin2(3x), we need (3x)2=9x2 in the denominator. Thus, we write:
sin2(3x)=(3xsin(3x))2⋅9x2
We apply the same logic to the other terms. For sin3(4x), we need (4x)3=64x3 in the denominator. For (loge(1+2x))5, we need (2x)5=32x5 in the denominator.
Phase 4
The Grand Cancellation
Now, let us assemble our pieces back into the limit:
x→0limcos3(4x)⋅(2xloge(1+2x))5⋅32x5(3xsin(3x))2⋅9x2⋅(4xsin(4x))3⋅64x3
Look at the powers of x. In the numerator, we have x2⋅x3=x5. In the denominator, we have x5. They cancel out perfectly.
Final Calculation
The indeterminate form vanishes, leaving us with constants. As x→0, all the standard limit fractions evaluate to 1.
We are left with:
1⋅1⋅321⋅9⋅1⋅64=329⋅64=9⋅2=18
And there it is—the elegance of mathematics in action. We started with a terrifying expression and ended with a simple, clean 18. Keep practicing this art of deconstruction, and no limit will ever scare you again.