Kindly solve the following limit.
$$\lim_{x \to 0} \frac{(x-\sqrt{2})^4}{\sin^3(x)\cos^2(x)}\int_{0}^x \frac{t^2}{t^4+1}dt$$
It's a $0/0$ form, so I thought of applying L'hospital rule but things got worse.
I can find no simplification. Hints please. I think we need to break that integral part though.
Hint:\begin{align}\lim_{x \to 0} \frac{(x-\sqrt{2})^4}{\sin^3(x)\cos^2(x)}\int_{0}^x \frac{t^2}{t^4+1}dt =\left(\lim_{x \to 0}\frac{(x-\sqrt2)^4x^3}{\cos^2x \sin^3(x)}\right)\left(\lim_{x \to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}\,dt\right)\end{align}
Evaluate the two limit on the right separately, use L'hopital rule on one of them.
We have that $$\begin{aligned} \lim_{x \to 0} \frac{(x-\sqrt{2})^4}{\sin^3(x)\cos^2(x)}\int_{0}^x \frac{t^2}{t^4+1}dt &= \left(\lim_{x \to 0} \frac{(x-\sqrt{2})^4}{\cos^2(x)}\right)\left(\lim_{x\to 0}\frac{\int_{0}^x \frac{t^2}{t^4+1}dt}{\sin^3(x)}\right) \\ &= 4\cdot \lim_{x\to 0}\frac{x^2}{3(x^4+1)\sin^2(x)\cos(x)} \\ &= \frac{4}{3}\cdot \lim_{x\to 0}\frac{x^2}{\sin^2(x)} \end{aligned}$$ where we are just pulling terms out of the product that can be evaluated using $\lim_{x\to x_0}f(x)/g(x)=f(x_0)/g(x_0)$ and applying L'Hospital's rule once to get rid of the integral.
[Not a complete solution but too long to be a comment.]
What troubles you when applying L'Hopital rule directly to $$ \lim_{x \to 0} \frac{(x-\sqrt{2})^4}{\sin^3(x)\cos^2(x)}\int_{0}^x \frac{t^2}{t^4+1}dt $$ is that there are some "noise functions" there making the derivatives complicated. But you could separate those parts out. Remember that one has the product rule $$ \lim_{x\to 0}g(x)f(x)=\lim_{x\to 0}g(x)\lim_{x\to 0}f(x) $$ when both $\lim_{x\to 0}g(x)$ and $\lim_{x\to 0}f(x)$ exist. One possible choice is to find $g(x)f(x)$ so that $$ \lim_{x\to 0}g(x)\neq 0 $$ and $$ \lim_{x\to 0}f(x) $$ could be calculated by L'Hopital rule with less complicated derivative-calculations.