Mathematics · 2019
JEE Main · 9 January 2019, Shift 1 · Q70
lim_y → 0 (√(1+√(1+y^4))-√ 2)/y^4
$$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}
$$
Official answer
(3)
exists and equals $\displaystyle \frac{1}{4 \sqrt{2}}$
More from Limits
- lim_x → 0(1+x)^(1/(e^x-1)) is equal to2021
- lim_x → ∞ ((√(3 x+1)+√(3 x-1))^6+(√(3 x+1)-√(3 x-1))^6)/((x+√(x^2-1))^6+(x-√(x^2-1))^6) x^32023
- The value of the limit: lim_x → 4 2^x+2^5-x-17^(1/(8-2^3 x / 4)) is equal to2021
- The value of lim_h → 0 2{(√3 sin ((π)/6+h)- cos ((π)/6+h))/(√3 h(√3 cos h- sin h))} is:2021
- The value of lim_x → 0 (log_e( sec (e x) · sec (e^2 x) · … · sec (e^10 x)))/(e^2-e^2 cos x) is equal to2026
- If lim_x → ∞((e/(1-e))(1/e-x/(1+x)))^x=α, then the value of (log_e α)/(1+ log_e α) equals:2025
- Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which lim_x →…2025
- If α is the positive root of the equation, p(x)=x^2-x-2=0, then lim_x → α^+ (√(1- cos (p(x))))/(x+α-4) is equal to:2020
JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.