Let [t] denote the greatest integer ≤ t. If for some λ ∈ R – {0,1}, lim(x→0) |1-x+|x|| / (λ−x+[x]) = L, then L is equal to:

$\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3k^2}{n^2}\right)}}$

$=\underset{n\to \infty }{lim}\frac{1}{n}{\displaystyle \sum _{k-1}^n\frac{1}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3k^2}{n^2}\right)}}$

$={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}}$

$={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{3}\times \frac{3}{2}\frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)}dx}$

$=\frac{1}{2}{\left[\sqrt[]{3}ta{n}^{-1}\left(\sqrt[]{3}x\right)\right]}_0^1-\frac{1}{2}{\left(ta{n}^{-1}x\right)}_0^1$

$=\frac{\sqrt[]{3}}{2}\left(\frac{\pi }{3}\right)-\frac{1}{2}\left(\frac{\pi }{4}\right)=\frac{\pi }{2\sqrt[]{3}}-\frac{\pi }{8}$

Lt? →? x/ (1−sinx)¹/? − (1+sinx)¹/? )
= 2x/ (1−sinx)¹/? − (1+sinx)¹/? ) Multiply by conjugate
= 4x/ (1−sinx)¹/²− (1+sinx)¹/²) Multiply by conjugate
= 8x/ (1−sinx−1−sinx) Multiply by conjugate
= 4x/sinx = −4

lim (x→0) [a e? - b cos (x) + c e? ] / (x sin (x) = 2
Using Taylor expansions around x=0:
lim (x→0) [a (1+x+x²/2!+.) - b (1-x²/2!+.) + c (1-x+x²/2!+.)] / (x * x) = 2
lim (x→0) [ (a-b+c) + x (a-c) + x² (a/2+b/2+c/2) + O (x³)] / x² = 2
For the limit to exist, the coefficients of lower powers of x in the numerator must be zero.
a - b + c = 0
a - c = 0 ⇒ a = c
Substituting a=c into the first equation: 2a - b = 0 ⇒ b = 2a.
The limit becomes: lim (x→0) [x² (a/2 + b/2 + c/2)] / x² = (a+b+c)/2
(a + b + c) / 2 = 2 ⇒ a + b + c = 4.

P (x) = 0
x² - x - 2 = 0
(x-2) (x+1) = 0
x = 2, -1 ∴ α = 2
Now lim (x→2? ) (√ (1-cos (x²-x-2) / (x-2)
⇒ lim (x→2? ) (√ (2sin² (x²-x-2)/2) / (x-2)
⇒ lim (x→2? ) (√2 sin (x²-x-2)/2) / (x²-x-2)/2) ⋅ (x²-x-2)/2) ⋅ (1/ (x-2)
⇒ for x→2? , (x²-x-2)/2 → 0?
⇒ lim (x→2? ) √2 ⋅ 1 ⋅ (x-2) (x+1)/ (2 (x-2) = 3/√2