Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2-\text{x}}-\sqrt{2+\text{x}}}{\text{x}}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{2-\text{x}}-\sqrt{2+\text{x}}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}{\text{x}\times\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(2-\text{x})-(2+\text{x})}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{-2\text{x}}{\text{x}\big(\sqrt{2-\text{x}}+\sqrt{2+\text{x}}\big)}$
$=\frac{-2}{\sqrt{2}+\sqrt{2}}$
$=\frac{-2}{2\sqrt{2}}$
$=\frac{-1}{\sqrt{2}}$
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$\sin(\text{x}+1)$
| Column A | Column B | ||
| a. | The polar form of $\text{i}+\sqrt{3}$ is | i. | Perpendicular bisector of segment joining (– 2, 0) and (2, 0). |
| b. | The amplitude of $-1+\sqrt{-3}$ is | ii. | On or outside the circle having centre at (0, – 4) and radius 3. |
| c. | If |z + 2| = |z - 2|, then locus of z is | iii. | $\frac{2\pi}{3}$ |
| d. | If |z + 2i| = |z - 2i|, then locus of z is | iv. | Perpendicular bisector of segment joining (0, – 2) and (0, 2). |
| e. | Region represented by $|\text{z}+4\text{i}|\geq3$ is | v. | $2\Big(\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$ |
| f. | Region represented by $|\text{z}+4|\leq3$ is | vi. | On or inside the circle having centre (– 4, 0) and radius 3 units. |
| g. | Conjugate of $\frac{1+2\text{i}}{1-\text{i}}$ lies in | vii. | First quadrant. |
| h. | Reciprocal of 1 - i lies in | viii. | Third quadrant. |
Using binomial theorem, prove that $2^{3\text{n}}-7\text{n}-1$ is divisible by 49 where $\text{n}\in\text{N}.$