下面是本人整理的一份九九乘法表，非常适合小学生学习。
1×1=limx→∞∫0xe−t dt1−e−x\lim_{x \to \infty} \frac{\int_0^x e^{-t} \, dt}{1-e^{-x}}limx→∞ 1−e−x∫0x e−tdt
1×2=limx→0x21−cosx\lim_{x \to 0} \frac{x^2}{1- \cos x}limx→0 1−cosxx2
1×3=limh→0∫01+h3x2 dx−∫013x2 dxh\lim_{h \to 0} \frac{\int_0^{1+h} 3x^2 \, dx - \int_0^1 3x^2 \, dx}{h}limh→0 h∫01+h 3x2dx−∫01 3x2dx
1×4=limn→∞4n∑k=1n−eiπ\lim_{n \to \infty} \frac{4}{n} \sum_{k=1}^{n} -e^{iπ}limn→∞ n4 ∑k=1n −eiπ
1×5=limn→∞∑k=1n[∫k−1nkn(limm→∞∑j=1m∫j−1mjm5 dt)dx]\lim_{n \to \infty}\sum_{k=1}^n\left[\int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(\lim_{m\to\infty}\sum_{j=1}^m\int_{\frac{j-1}{m}}^{\frac{j}{m}}5 \, dt\right) dx\right]limn→∞ ∑k=1n [∫nk−1 nk (limm→∞ ∑j=1m ∫mj−1 mj 5dt)dx]
1×6=limn→∞∑k=1n[∫k−1nkn(6n⋅n2π⋅e−n2(x−kn)2)dx]×1n\lim_{n \to \infty}\sum_{k=1}^n\left[\int_{\frac{k-1}{n}}^{\frac{k}{n}}\left(6n \cdot \sqrt{\frac{n}{2\pi}} \cdot e^{-\frac{n}{2}(x - \frac{k}{n})^2}\right) dx\right] \times \frac{1}{\sqrt{n}}limn→∞ ∑k=1n [∫nk−1 nk (6n⋅2πn ⋅e−2n (x−nk )2)dx]×n 1
1×7=limn→∞∫01[∑k=1n14ncos(2πkx)+7n]dx⋅n\lim_{n \to \infty}\int_0^1\left[\sum_{k=1}^n\frac{14}{n} \cos(2\pi k x)+ \frac{7}{n}\right] dx \cdot nlimn→∞ ∫01 [∑k=1n n14 cos(2πkx)+n7 ]dx⋅n
1×8=limn→∞∑k=1n[∫01(8n∑j=0n(nj)xj(1−x)n−j)⋅(1Q(x)−eiπ)dx]\lim_{n \to \infty}\sum_{k=1}^n\left[\int_0^1\left(\frac{8}{n}\sum_{j=0}^n\binom{n}{j}x^j (1-x)^{n-j}\right)\cdot\left(\mathbf{1}_{\mathbb{Q}}(x)-e^{iπ}\right)dx\right]limn→∞ ∑k=1n [∫01 (n8 ∑j=0n (jn )xj(1−x)n−j)⋅(1Q (x)−eiπ)dx]
1×9=limn→∞[nπ∑k=−n2n2e−(k/n)2⋅1n]⋅[9nπ∑k=−n2n2e−(k/n)2⋅1n]\lim_{n \to \infty} \left[ \frac{n}{\sqrt{\pi}} \sum_{k=-n^2}^{n^2} e^{-(k/n)^2} \cdot \frac{1}{n} \right] \cdot \left[ \frac{9n}{\sqrt{\pi}} \sum_{k=-n^2}^{n^2} e^{-(k/n)^2} \cdot \frac{1}{n} \right]limn→∞ [π n ∑k=−n2n2 e−(k/n)2⋅n1 ]⋅[π 9n
∑k=−n2n2 e−(k/n)2⋅n1 ]
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