Answer
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^1}=0$
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^2}=1$
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^{2n+1}}$ does not exist
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^{2n}}=\infty$
Work Step by Step
We have to determine:
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^k}$
Compute the limit for $k=1$:
$\dfrac{\sin (\sin^2 (-0.01))}{(-0.01)^1}\approx -0.0099996667$
$\dfrac{\sin (\sin^2 (-0.001))}{(-0.001)^1}\approx -0.0009999997$
$\dfrac{\sin (\sin^2 (-0.0001))}{(-0.0001)^1}\approx -0.0001$
$\dfrac{\sin (\sin^2 (-0.00001))}{(-0.00001)^1}\approx -0.00001$
$\dfrac{\sin (\sin^2 0.00001)}{0.00001^1}\approx 0.00001$
$\dfrac{\sin (\sin^2 0.0001)}{0.0001^1}\approx 0.0001$
$\dfrac{\sin (\sin^2 0.001)}{0.001^1}\approx 0.0009999997$
$\dfrac{\sin (\sin^2 0.1)}{0.1^1}\approx 0.0099996667$
Therefore we got:
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^1}=0$
Compute the limit for $k=2$:
$\dfrac{\sin (\sin^2 (-0.01))}{(-0.01)^2}\approx 0.99996667$
$\dfrac{\sin (\sin^2 (-0.001))}{(-0.001)^2}\approx 0.99999967$
$\dfrac{\sin (\sin^2 (-0.0001))}{(-0.0001)^2}\approx 1$
$\dfrac{\sin (\sin^2 (-0.00001))}{(-0.00001)^2}\approx 1$
$\dfrac{\sin (\sin^2 0.00001)}{0.00001^2}\approx 1$
$\dfrac{\sin (\sin^2 0.0001)}{0.0001^2}\approx 1$
$\dfrac{\sin (\sin^2 0.001)}{0.001^2}\approx 0.99999967$
$\dfrac{\sin (\sin^2 0.1)}{0.1^2}\approx 0.99996667$
Therefore we got:
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^2}=1$
Compute the limit for $k=3$:
$\dfrac{\sin (\sin^2 (-0.01))}{(-0.01)^3}\approx -99.996667$
$\dfrac{\sin (\sin^2 (-0.001))}{(-0.001)^3}\approx -999.99967$
$\dfrac{\sin (\sin^2 (-0.0001))}{(-0.0001)^3}\approx -10,000$
$\dfrac{\sin (\sin^2 (-0.00001))}{(-0.00001)^3}\approx -100,000$
$\dfrac{\sin (\sin^2 0.00001)}{0.00001^3}\approx 100,000$
$\dfrac{\sin (\sin^2 0.0001)}{0.0001^3}\approx 10,000$
$\dfrac{\sin (\sin^2 0.001)}{0.001^3}\approx 999.99967$
$\dfrac{\sin (\sin^2 0.1)}{0.1^3}\approx 99.996667$
Therefore we got:
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^3}$ does not exist
Compute the limit for $k=4$:
$\dfrac{\sin (\sin^2 (-0.01))}{(-0.01)^4}\approx 9999.6667$
$\dfrac{\sin (\sin^2 (-0.001))}{(-0.001)^4}\approx 999,999.67$
$\dfrac{\sin (\sin^2 (-0.0001))}{(-0.0001)^4}\approx 10^8$
$\dfrac{\sin (\sin^2 (-0.00001))}{(-0.00001)^4}\approx 10^{10}$
$\dfrac{\sin (\sin^2 0.00001)}{0.00001^4}\approx 10^{10}$
$\dfrac{\sin (\sin^2 0.0001)}{0.0001^4}\approx 10^8$
$\dfrac{\sin (\sin^2 0.001)}{0.001^4}\approx 999,999.67$
$\dfrac{\sin (\sin^2 0.1)}{0.1^4}\approx 9999.6667$
Therefore we got:
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^4}=\infty$
Therefore we got:
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^1}=0$
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^2}=1$
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^{2n+1}}$ does not exist
$\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin (\sin^2 x)}{x^{2n}}=\infty$
