Let's say I have the expression $-b+\frac{3}{2}b\cdot a^2$. Can I say that by taking the limit when $a<<1$, that expression is $\approx -b$ ? can a constant number, like $\frac{3}{2}$ can "ruin" the limit?
Thanks.
2026-04-05 17:47:43.1775411263
Boundary of the limit when $a<<1$
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The answer might be yes. You may observe that $$ -b+\frac{3}{2}b\cdot a^2=-b\left(1-\frac{3a^2}{2}\right) $$ giving, as $a \to 0$, $$ -b+\frac{3}{2}b\cdot a^2\approx-b $$ since$$ \left(1-\frac{3a^2}{2}\right) \to 1.$$