Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-64}{\text{x}^2-16}$
$\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-64}{\text{x}^2-16}$
$=\lim\limits_{\text{x}\rightarrow{4}}\frac{\text{x}^{3}-4^3}{\text{x}^2-4^2}$
$=\lim\limits_{\text{x}\rightarrow{4}}\frac{\frac{\text{x}^{3}-4^3}{\text{x}-4}}{\frac{\text{x}^2-4^2}{\text{x}-4}}$ [Dividing numerator and denominator by x - 4]
$=\frac{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^3-4^3}{\text{x}-4}}{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-4}{\text{x}-4}}$
Applying formula $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{n}}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na}^{\text{n}-1}$ in numerator and $\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{x}^{\text{m}}-\text{a}^{\text{m}}}{\text{x}-\text{a}}=\text{ma}^{\text{m}-1}$ in denominator
$\Rightarrow\text{n}=3,\text{m}=2$
$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^3-4^3}{\text{x}-4}}{\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-4}{\text{x}-4}}=\frac{3(4)^{3-1}}{2(4)^{2-1}}=\frac{3(4)^2}{2(4)}$
$=6$
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Evaluate the following:
$\Big(\text{x}+\sqrt{\text{x}^2-1}\Big)^6+\Big(\text{x}-\sqrt{\text{x}^2-1}\Big)^6$
$25\text{x}^2+16\text{y}^2=1600.$
| xi | $1\leq\text{x}<3$ | $3\leq\text{x}<5$ | $5\leq\text{x}<7$ | $7\leq\text{x}<10$ |
| f1 | 6 | 4 | 5 | 1 |
Using binomial evaluate the following:
$(98)^5$