JEE Main 2020 · 5 September · evening · Single correct · +4 / −1
lim _(x → 0) x( e^(( √(1 + x² + x⁴) - 1 )/x) - 1 )/(√(1 + x² + x⁴) - 1)
- (a) is equal to 0.
- (b) is equal to √e.
- (c) is equal to 1.
- (d) does not exist.
Answer and solution
Answer (c) is equal to 1.
Solution
Maths sir · 4 steps
Call the exponent t; the whole expression becomes (eᵗ-1)/t, and t goes to 0.
- Name the exponent. Let t = (√(1+x²+x⁴)-1)/x. Then the denominator √(1+x²+x⁴)-1 is exactly xt.
x(eᵗ-1)/xt = (eᵗ-1)/t
- Find where t goes. Rationalise: t = (x²+x⁴)/x(√(1+x²+x⁴)+1) = x(1+x²)/(√(1+x²+x⁴)+1). As x → 0 this tends to 0/2 = 0, from both sides.
- Use the standard limit. Since t → 0, the standard result lim(t→ 0)(eᵗ-1)/t = 1 applies directly.
- So. The limit is 1, option (c).
The idea to keep. When you see e⁽ˢᵒᵐᵉᵗʰⁱⁿᵍ⁾-1, check whether that same something sits in the denominator. If it does and it tends to 0, the limit is 1.
Answer from the NTA final answer key. Solution by PrepWiser.