===ג===
<math>L=\frac{L^2}{2}+\frac{1}{2}</math> ולכן <math>L=1</math>
===ד===
<math>\Big(1+\frac{3n}{n^2+1}\Big)^n
=\Big(1+\frac{1}{\frac{n}{3}+\frac{1}{3n}}\Big)^{n\cdot\frac{\frac{n}{3}+\frac{1}{3n}}{\frac{n}{3}+\frac{1}{3n}}}
=\Big(1+\frac{1}{\frac{n}{3}+\frac{1}{3n}}\Big)^{(\frac{n}{3}+\frac{1}{3n})\cdot{\frac{n}{\frac{n}{3}+\frac{1}{3n}}}}
\rightarrow e^3
</math>