Limits and Derivatives

LHL : lim_ (x→0? ) |1-x-x|/|λ-x-1| = 1/|λ-1|
RHL: lim_ (x→0? ) |1-x+x|/|λ-x+0| = 1/|λ|
For existence of limit
LHL = RHL
⇒ 1/|λ-1| = 1/|λ| ⇒ λ = 1/2
∴ L = 1/|λ| = 2

lim (x?0) (tan (? /4+x)¹/? = e^ (lim (x?0) (tan (? /4+x)-1)/x)
= e^ (lim (x?0) (2tanx/ (1-tanx)/x) = e².

lim (x→1) (x+x²+.+x? -n)/ (x-1) = 820
⇒ lim (x→1) (x-1)/ (x-1)+ (x²-1)/ (x-1)+.+ (x? -1)/ (x-1) = 820
⇒ 1+2+.+n = 820
⇒ n (n+1)=2*820
⇒ n (n+1)=40*41
Since n∈N, so n=40

Determinant = -2-sin2x. Max -1, min -3. (m, M)= (-3, -1).

Let x = tanθ
y? = tan? ¹ (secθ-1)/tanθ) = tan? ¹ (tan (θ/2) = θ/2 = (1/2)tan? ¹x
x = sinφ, y? = tan? ¹ (2sinφcosφ)/cos2φ)
= tan? ¹ (tan2φ) = 2φ = 2sin? ¹x
dy? /dy? = (dy? /dx)/ (dy? /dx) = (1/2) (1/ (1+x²) / (2/√ (1-x²)
= (√ (1-x²) / (4 (1+x²) = (√ (1-1/4) / (4 (1+1/4) = √3 / 10

lim (x→0) (e^ (√ (1+x²+x? )-1)/x - 1) / ( (√ (1+x²+x? )-1)/x )
put (√ (1+x²+x? )-1)/x = t
clearly x→0 ⇒ t→0
∴ given limit = lim (t→0) (e? -1)/t = 1

Let lim (x→0) [ (3x²+2)/ (7x²+2)]^ (1/x²) = e^ [lim (x→0) (1/x²) * (3x²+2)/ (7x²+2) - 1)] =

e^ [lim (x→0) (1/x²) * (-4x²)/ (7x²+2)] = e^ (-4/2) = e? ²

$\mathrm{}{f}^{\mathrm{\text{'}}}\left(x\right)=a(x+1)(x-1)x^2$

$f^{\mathrm{\text{'}}}\left(x\right)=a{x}^4-x^{2}f$

$f\left(x\right)=\frac{a{x}^5}{5}-\frac{a{x}^3}{5}+C$
$?f\left(0\right)=0\Rightarrow c=0$

${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x^3}=2$

$\Rightarrow \mathrm{a}=-6$

Minima at $\therefore f^{\mathrm{\text{'}}}\left(x\right)=-6\left(x^2-1\right)\left(x^2\right)$

Maxima at $x=-1$

 $\underset{x\to 0}{lim}\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)}=b,$ use of L' Hospital rule implies

$\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a\left(e^{4x}-1\right)+ax\left(4e^{4x}\right)}$

$=\frac{a-4}{0}\Rightarrow a=4$

$\Rightarrow \underset{n\text{?}\to 0}{lim}\frac{4\left(-4.e^{4x}\right)}{4.4e^{4x}+16e^{4x}+16x.4e^{4x}}$

$=\frac{-16}{16+16}=-\frac{1}{2}=b$

a – 2b = 4 – (1) = 5

f : R -> R.

$f\left(x\right)=[\begin{array}{l}\frac{x^3}{{\left(1-cos2x\right)}^2}log\left(\frac{1+2x{e}^{-2x}}{{\left(1-x{e}^{-x}\right)}^2}\right)x\ne 0\\ \alpha x=0\end{array}$                

As f is continuous at x = 0

$\therefore \alpha =\underset{x\to 0}{Lim}f\left(x\right)$              

$=\underset{x\to 0}{Lim}\frac{1}{2}\left[e^{-2x}+e^{-x}\right]=\frac{1}{2}*2=1$            

$\therefore \alpha =1$