Find $\,\lim\limits_{t\to0^+}t\!\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x$

Question:

Find $\;\displaystyle\lim_{t\to0^+}\,t\!\!\int_{t^2}^t\frac{\cos(x)}{x^{3/2}}\,\mathrm{d}x$.

Attempted solution:

Let: $$f(x)=\begin{cases}\frac{\cos x-1}{x^{3/2}}&\text{if }x\in(0,\pi/2)\\0&\text{otherwise}\end{cases}$$

$\lim\limits_{x\to0^+}f(x)=0\;$ so $\,f(x)\,$ is continuous on $[0,\pi/2]$.

So, $\;\displaystyle\left|\int_{t^2}^t\frac{\cos x-1}{x^{3/2}}\,\mathrm{d}x\right|\leq\int_0^t \frac{1-\cos x}{x^{3/2}}\,\mathrm{d}x\;,$
for $t\in[0,\pi/2]$, which tends to zero as $t\to0^+$.

$$\begin{align}t\int_{t^2}^t\frac{\cos x}{x^{3/2}}\,\mathrm{d}x&=t\left[\int_{t^2}^t\frac{\cos x-1}{x^{3/2}}\,\mathrm{d}x+\int_{t^2}^t\frac{1}{x^{3/2}}\,\mathrm{d}x\right]\\&=O(t)+2 (1-\sqrt{t})\end{align}$$

So the answer is $2$.

I just wanted to make sure is my approach correct.