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Automated Insulin Delivery Infusion (Biomedical Engineering) Solution

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Problem Overview

We evaluate the right-sided limit of the residual insulin concentration function:

I(t)=101+e−1/t2I(t) = \frac{10}{1 + e^{-1/t^2}}
as time approaches zero from the positive side (t→0+t \to 0^+).

Step-by-Step Mathematical Solution

Step 1: Evaluate the innermost exponent fraction

Look at the nested rational term u(t)=−1t2u(t) = -\frac{1}{t^2} as t→0+t \to 0^+: As tt approaches 00 through positive values, t2t^2 approaches 00 through positive values (t2→0+t^2 \to 0^+). Therefore:

lim⁡t→0+1t2=+∞  ⟹  lim⁡t→0+(−1t2)=−∞\lim_{t \to 0^+} \frac{1}{t^2} = +\infty \implies \lim_{t \to 0^+} \left(-\frac{1}{t^2}\right) = -\infty

Step 2: Evaluate the exponential intermediate layer

Let u=−1t2u = -\frac{1}{t^2}. As u→−∞u \to -\infty, evaluate the natural exponential decay:

lim⁡u→−∞eu=0\lim_{u \to -\infty} e^u = 0

Step 3: Evaluate the overall rational composite expression

Substitute the evaluated exponential limit (00) back into the denominator of the full expression:

lim⁡t→0+101+e−1/t2=101+lim⁡t→0+e−1/t2=101+0=101=10 Units\lim_{t \to 0^+} \frac{10}{1 + e^{-1/t^2}} = \frac{10}{1 + \lim_{t \to 0^+} e^{-1/t^2}} = \frac{10}{1 + 0} = \frac{10}{1} = 10\text{ Units}

Engineering Conclusion

The composite limit cleanly evaluates to lim⁡t→0+I(t)=10 Units\lim_{t \to 0^+} I(t) = 10\text{ Units}. This confirms that the infusion algorithm reliably initializes at the full basal dosage of 10 Units10\text{ Units} at the moment of valve opening (t→0+t \to 0^+), eliminating mathematical numerical instabilities and preventing a fatal insulin overdose.