The value of limₓ→₀ x/(√1-sinx - √1+sinx) is equal to :

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

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

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

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

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

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

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 n

Let t = 3^ (x/2). As x→2, t→3^ (2/2) = 3.
The limit becomes lim (t→3) [ (t² + 27/t²) - 12 ] / [ (t - 3²/t) ].
lim (t→3) [ (t? - 12t² + 27)/t² ] / [ (t² - 9)/t ].
lim (t→3) [ (t²-9) (t²-3) / t² ] * [ t / (t²-9) ].
lim (t→3) [ (t²-3) / t ].
Substituting t=3: (3²-3)/3 = (9-3)/3 = 6.
(The provided solution a

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

Applying L'Hôpital's Rule
Lim (t→x) [2tf² (x) – x² (2f (t)f' (t)] / 1
∴ 2xf² (x) – x² (2f (x)f' (x) = 0
⇒ f (x) – xf' (x) = 0
⇒ f' (x)/f (x) = 1/x ⇒ lnf (x) = lnx + C
At x=1, c=1
∴ lnf (x) = lnx + 1
when f (x) = 1
then lnx = -1
x = 1/e