Limit of an integral in the form of $\lim_{x\to 0} g(x) \int_{0}^{x} f(t) dt$

Since $\sinh(x)\sim x$ and $\cos(x)-1\sim -\frac{x^2}2$ as $x\to 0$ we get $$\frac{\sinh(x)}{\cos(x)-1} \int_{0}^{x} \sqrt{e^t-t^4}\mathrm dt\sim-\frac2{x}\int_{0}^{x} \sqrt{e^t-t^4}\mathrm dt$$ De L'Hopital rule reduces it to $$-2\sqrt{e^x-x^4}\xrightarrow{x\to 0}-2$$

What you did is right, but your first thought (L'Hospital) would be a quicker way to do this problem. If you let $$f(x)=\int_0^x\sqrt{e^x-x^4}dx,$$ then the limit is $$\lim_{h\rightarrow 0} \frac{\sinh (x)f(x)}{\cos (x)-1},$$ and since $f'(x)=\sqrt{e^x-x^4}$, using L'Hospital twice will give you the answer

The problem is easily handled by rewriting the expression under limit as $$-\dfrac{x^2} {\sin^2 x}\cdot\frac{\sinh x} {x} \cdot (\cos x +1)\cdot\frac{1}{x}\int_{0}^{x}f(t)\,dt$$ where $f(t) =\sqrt{e^t-t^4}$. Now the limit of first fraction is $1$, that of second fraction is $1$, that of third factor is $2$ and the last factor tends to $f(0)=1$ via Fundamental Theorem of Calculus. The desired answer is thus $-2$.

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If you see an integral of type $\int_{0}^{x}f(t)\,dt $ in limit evaluation as $x\to 0$, it makes sense to rewrite it as $$x\cdot\frac{1}{x}\int_{0}^{x}f(t)\,dt$$ and then one uses FTC.