Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of limx→0(81−sinx−81+sinxx) is equal to:
Select Answer:
Visualized Solution
limx→081−sinx−81+sinxx
Given expression: limx→0(81−sinx−81+sinxx)
Goal: Evaluate the limit as x approaches 0.
Observe the structure: A linear numerator and a difference of roots in the denominator.
Indeterminate Form 00
Check for indeterminate form by substituting x=0.
Numerator: x=0.
Denominator: 81−sin(0)−81+sin(0)=81−81=1−1=0.
The limit is in the 00 form.
Binomial Approximation Tool
Recall Binomial Approximation: (1+t)n≈1+nt for ∣t∣≪1.
As x→0, sinx→0. Thus, t=±sinx is a small quantity.
This tool will help us linearize the radical expressions in the denominator.
Rewriting as Fractional Powers
Rewrite the radicals using fractional exponents:
81−sinx=(1−sinx)81
81+sinx=(1+sinx)81
The denominator is now (1−sinx)81−(1+sinx)81.
Approximating (1−sinx)81
Apply (1+t)n≈1+nt to the first term.
Here t=−sinx and n=81.
(1−sinx)81≈1+81(−sinx)=1−81sinx.
Approximating (1+sinx)81
Apply (1+t)n≈1+nt to the second term.
Here t=sinx and n=81.
(1+sinx)81≈1+81(sinx)=1+81sinx.
Substituting Approximations
Substitute the approximations back into the denominator of the limit:
Denominator ≈(1−81sinx)−(1+81sinx)
Simplifying the Denominator
Simplify the expression:
(1−81sinx)−1−81sinx
=−81sinx−81sinx=−82sinx
=−41sinx
Rearranging the Limit
The limit becomes: limx→0−41sinxx
Factor out the constant: −4⋅limx→0sinxx
Standard Limit limx→0sinxx
Using the standard limit: limx→0xsinx=1⟹limx→0sinxx=1.
Substitute the value: −4⋅(1)=−4.
The final value of the limit is −4.
Key Takeaways
Key Concept: Binomial Approximation (1+t)n≈1+nt is powerful for limits involving roots as t→0.
Standard Limit: Always look to reduce expressions to limx→0xsinx=1.
Next Challenge: Try solving the same problem if the roots were of order n instead of 8. Does the answer become −2n or something else?
00:00 / 00:00
The Sigma Insight: Evaluation of Limits & L'Hopital's Rule
Solution Diagram
The Anatomy of a Limit
Facing the Eighth Root Monster
Welcome, future engineer. Today we are going to dismantle a problem that, at first glance, looks like a mathematical nightmare. We are looking at the limit:
x→0lim81−sinx−81+sinxx
Eighth roots? It sounds terrifying, doesn't it? But here is the secret: in the world of JEE Advanced, intimidation is just a test of your conceptual clarity. Let's break this down together.
Phase 1
The Diagnostic
Before we panic, let's perform a simple diagnostic. What happens when x approaches 0?
The numerator becomes 0. The denominator becomes 81−sin(0)−81+sin(0), which is 81−81=0.
We have a 0/0 indeterminate form. This is our green light; it tells us that there is a hidden structure waiting to be revealed. We don't need to run away; we just need to simplify.
Phase 2
The Weaponry
How do we handle these eighth roots? We use the Binomial Approximation. Recall that for a small quantity t, (1+t)n≈1+nt.
This is the scalpel that will cut through the complexity. Since x→0, sinx is also approaching 0. This makes sinx our tiny quantity t.
We can rewrite our denominator using fractional exponents:
(1−sinx)1/8−(1+sinx)1/8
Now, we apply the approximation to each term. For the first term, t=−sinx and n=1/8:
(1−sinx)1/8≈1+81(−sinx)=1−81sinx
For the second term, t=sinx and n=1/8:
(1+sinx)1/8≈1+81(sinx)=1+81sinx
Phase 3
The Transformation
Now, watch the magic happen. We substitute these approximations back into our denominator:
(1−81sinx)−(1+81sinx)
Let's simplify this carefully. The 1 and −1 cancel out perfectly, leaving us with:
−81sinx−81sinx=−82sinx=−41sinx
The complex roots have vanished, replaced by a simple trigonometric term.
Phase 4
The Final Act
Our limit now looks like this:
x→0lim−41sinxx
We can pull the constant −4 out of the limit:
−4⋅x→0limsinxx
We know the standard limit limx→0xsinx=1, which means its reciprocal limx→0sinxx is also 1. Therefore, our final answer is:
−4⋅1=−4
See? The monster was just a paper tiger. By staying calm and applying the right tools, you turned a terrifying expression into a simple, elegant result. Keep this mindset for every problem you face, and you will conquer the JEE.