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Prove that $4e_{x}$ is equal to the sum of its Maclaurin series. Solution If $f(x)=4e_{x}$, then $f_{(n+1)}(x)=$ for all $n$. If $d$ is any positive number and $∣x∣≤d,then∣∣ f_{(n+1)}(x)∣∣ =$ So Taylor's Inequality, with $a=0$ and $M=4e_{d}$, says that $∣R_{n}(x)∣≤(n+1)!∣x∣_{n+1} for∣x∣≤d.$ Notice that the same constant $M=4e_{d}$ works for every value of $n$. But, from this equation, we have $lim_{n→∞}(n+1)!−4e_{d} ∣x∣_{n+1}=4e_{d}lim_{n→∞}(n+1)!∣x∣_{n+1} =$ It follows from the Squeeze Theorem that $lim_{n→∞}∣R_{n}(x)∣=0$ and therefore $lim_{n→∞}R_{n}(x)=$ for ail values of $x$. By this theorem, Ae $e_{x}$ is equal to the sum of its Maclaurin series, that is, $4e_{x}=∑_{n=0}n!4x_{n} for allx.$

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