Solve the Following Question.(4 Marks)

Question 1014 Marks

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$

Answer

$\lim\limits_{\text{x}\rightarrow{1}}\frac{1-\frac{1}{\text{x}}}{\sin\pi(\text{x}-1)}$
As x → 1, then x - 1 → 0 let x - 1 = y
$=\lim\limits_{{\text{x}-{1\rightarrow0}}}\frac{(\text{x}-1)}{\text{x}\times\sin\pi(\text{x}-1)}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{\text{y}}{(\text{y}+1)\sin(\pi\text{y})}$
$=\lim\limits_{{\text{y}\rightarrow0}}\frac{1}{\frac{(\text{y}+1)\sin(\pi\text{y})}{\text{y}}}$
$=\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow0}(\text{y}+1)\Big)\times\Big(\lim\limits_{\text{y}\rightarrow0}\frac{\sin\pi\text{y}}{\text{y}\times\pi}\times\pi\Big)}$
$=\frac{1}{(1)(1\times\pi)}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{1}{\pi}$

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Question 1024 Marks

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{2}+1\big)}{\text{x}^2-2}$

Answer

$\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{2}+1\big)}{\text{x}^2-2}$$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\Big(\sqrt{\big(\sqrt{2}+1\big)^2}\Big)}{\text{x}^2-2}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\big(\sqrt{3+2\sqrt{2}}\big)}{\text{x}^2-2}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{3+2\text{x}}-\big(\sqrt{3+2\sqrt{2}}\big)}{\text{x}^2-2}\times\frac{\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)}{\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{{3+2\text{x}}-{3-2\sqrt{2}}}{\big(\text{x}^2-2\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{2\big(\text{x}-\sqrt{2}\big)}{\big(\text{x}^2-2\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{2}{\big(\text{x}^2-\sqrt{2}\big)\Big(\sqrt{3+2\text{x}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\frac{2}{\big(\sqrt{2}+\sqrt{2}\big)\Big(\sqrt{3+2\sqrt{2}}+\big(\sqrt{3+2\sqrt{2}}\big)\Big)}$
$=\frac{2}{\big(2\sqrt{2}\big)\Big(2\sqrt{3+2\sqrt{2}}\Big)}$
$=\frac{1}{\big(2\sqrt{2}\big)\Big(\sqrt{3+2\sqrt{2}}\Big)}$
$=\frac{1}{\big(2\sqrt{2}\big)\Big(\sqrt{2}+1\Big)}$
$=\frac{\big(\sqrt{2}-1\big)}{\big(2\sqrt{2}\big)}$

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Question 1034 Marks

Evaluate the following limit:
$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \cdots+\text{n}^2}{\text{n}^3}$

Answer

$\lim\limits_{\text{n}\rightarrow\infty}\frac{1^2+2^2+\ \cdots+\text{n}^2}{\text{n}^3}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac16\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{\text{n}^3}$ $\Big[1^2+2^2+3^2+\ \cdots+\text{n}^2=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+1)\Big]$
$=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(\text{n}^2+\text{n}\big)(2\text{n}+1)}{\text{n}^3}$
$=\frac{1}{6}\lim\limits_{\text{n}\rightarrow\infty}\frac{\big(2\text{n}^3+\text{n}^2+2\text{n}^2+\text{n}\big)}{\text{n}^3}$
$=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\big(2\text{n}^2+3\text{n}^2+\text{n}\big)}{\text{n}^3}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\frac16\lim\limits_{\text{n}\rightarrow{\infty}}\frac{\Big(2+\frac{3}{\text{n}}+\frac{1}{\text{n}^2}\Big)}{1}$
$=\frac16\frac{(2+0+0)}{1}=\frac{1}{3}$

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Question 1044 Marks

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2\tan2\text{x}}{\tan\text{x}}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2\tan2\text{x}}{\tan\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\big(\frac{\text{x}^2}{2\text{x}}-\frac{\tan2\text{x}}{2\text{x}}\big)\times2\text{x}}{\frac{\tan\text{x}}{\text{x}}\times\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\big(\frac{\text{x}}{2}-\frac{\tan2\text{x}}{2\text{x}}\big)}{\frac{\tan\text{x}}{\text{x}}}$
$=2\Big(\frac{0-1}{1}\Big)$
$=-2$

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Question 1054 Marks

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}$

Answer

$\lim\limits_{\text{x}\rightarrow\infty}\sqrt{\text{x}}\Big\{\sqrt{\text{x}+1}-\sqrt{\text{x}}\Big\}$$=\lim\limits_{\text{x}\rightarrow\infty}\Big(\sqrt{\text{x}^2+\text{x}}-{\text{x}}\Big)$
$=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\big(\sqrt{\text{x}^2+\text{x}}-\text{x}\big)\times\frac{\big(\sqrt{\text{x}^2+\text{x}}+\text{x}\big)}{\sqrt{\text{x}^2+\text{x}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{n}\rightarrow\infty}\Bigg(\frac{\big(\text{x}^2+\text{x}\big)-\text{x}^2}{\sqrt{\text{x}^2+\text{x}}+\text{x}}\Bigg)$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg(\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}}+\text{x}}\bigg)$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\begin{pmatrix}\frac{1}{\sqrt{\frac{\text{x}^2}{\text{x}^2}}+\frac{\text{x}}{\text{x}^2}+1}\end{pmatrix}$
$=\lim\limits_{\text{n}\rightarrow{\infty}}\begin{pmatrix}\frac{1}{\sqrt{1+\frac{1}{\text{x}}}+1}\end{pmatrix}$
$=\frac{1}{1+1}$
$=\frac12$

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Maths Solve the Following Question.(4 Marks)