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How to find this limit?

$$ I = \lim_{t \to 0+0} \int_1^3 \frac{\sin(tx)}{t} \sqrt{x^2 + t^2 + tx + 1} dx $$

The problem is that it is impossible to make the limit transition $$ I = \int_1^3 \lim_{t \to 0+0} \frac{\sin(tx)}{t} \sqrt{x^2 + t^2 + tx + 1} dx, $$ because the integrand function has a discontinuity in $t = 0$, therefore, the previous statement is incorrect, generally speaking.

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  • $\begingroup$ what do you mean from $t \rightarrow 0+0 $? $\endgroup$ – Khosrotash Feb 10 '15 at 17:56
  • $\begingroup$ @darya-khosrotash, one-sided limit, $t \to 0$ and $t > 0$. $\endgroup$ – Nastya Koroleva Feb 10 '15 at 17:59
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$$ \frac{\sin(t\,x)}{t}=x\,\frac{\sin(t\,x)}{t\,x}=x\,\text{sinc}(t\,x), $$ and the function $$ \text{sinc}(z)=\begin{cases}\dfrac{\sin z}{z} & z\ne0\\ 1 & z=0 \end{cases} $$ is continuous.

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