Calculus - Limits

Limit ပုစ္ဆာများ တွက်ရာတွင် သိရှိနားလည်ထားရမည့် ဂုဏ်သတ္တိများ ကို လေ့လာသိရှိပြီးမှ ပုစ္ဆာများကို တွက်လျှင် ပို၍ နားလည်လွယ်ကူစေပါသည်။


1.      $ \displaystyle \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}$

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$ \displaystyle \ \ \ \ \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}=\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(x-1)({{x}^{2}}+x+1)}}{{x-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left[ {{{x}^{2}}+x+1} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {{{x}^{2}}} \right)+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( x \right)+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( 1 \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3$


2.      $ \displaystyle \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x-1}}{{\sqrt[3]{x}-1}}$

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$ \displaystyle \ \ \ \ \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x-1}}{{\sqrt[3]{x}-1}}\ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{{\left( {\sqrt[3]{x}} \right)}}^{3}}-{{1}^{3}}}}{{\sqrt[3]{x}-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(\sqrt[3]{x}-1)\left[ {{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\sqrt[3]{x}+1} \right]}}{{\sqrt[3]{x}-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left[ {{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\sqrt[3]{x}+1} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,{{\left( {\sqrt[3]{x}} \right)}^{2}}+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {\sqrt[3]{x}} \right)+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( 1 \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3$


3.      $ \displaystyle \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}-9}}{{{{x}^{2}}+\sqrt{3}x-6}}\ $

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$ \displaystyle \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}-9}}{{{{x}^{2}}+\sqrt{3}x-6}}\ \ =\underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{({{x}^{2}}-3)({{x}^{2}}+3)}}{{(x-\sqrt{3})(x+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{(x-\sqrt{3})(x+\sqrt{3})({{x}^{2}}+3)}}{{(x-\sqrt{3})(x+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{(x+\sqrt{3})({{x}^{2}}+3)}}{{(x+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{{(\sqrt{3}+\sqrt{3})(3+3)}}{{(\sqrt{3}+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{{12\sqrt{3}}}{{3\sqrt{3}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 4$


4.      $ \displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{{{x}^{2}}-x-6}}\ $

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$ \displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{{{x}^{2}}-x-6}}\ =\underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{(x+2)({{x}^{2}}+5)}}{{(x+2)(x-3)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+5}}{{x-3}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{4+5}}{{-2-3}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{9}{5}$


5.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{3{{x}^{3}}-x-6}}\ $

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$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{3{{x}^{3}}-x-6}}\ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}\left( {1+\frac{2}{x}+\frac{5}{{{{x}^{2}}}}+\frac{{10}}{{{{x}^{3}}}}} \right)}}{{{{x}^{3}}\left( {3-\frac{1}{{{{x}^{2}}}}-\frac{6}{{{{x}^{3}}}}} \right)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{1+\frac{2}{x}+\frac{5}{{{{x}^{2}}}}+\frac{{10}}{{{{x}^{3}}}}}}{{3-\frac{1}{{{{x}^{2}}}}-\frac{6}{{{{x}^{3}}}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{1+0+0+0}}{{3-0-0}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{3}$


6.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{{-3}}}+5{{x}^{{-2}}}+8}}{{5{{x}^{{-3}}}-{{x}^{{-1}}}-2}}\ $

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$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{{-3}}}+5{{x}^{{-2}}}+8}}{{5{{x}^{{-3}}}-{{x}^{{-1}}}-2}}\ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+\frac{5}{{{{x}^{2}}}}+8}}{{\frac{5}{{{{x}^{3}}}}-\frac{1}{x}-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+\frac{5}{{{{x}^{2}}}}+8}}{{\frac{5}{{{{x}^{3}}}}-\frac{1}{x}-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{0+0+8}}{{0-0-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-4$


7.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2{{x}^{4}}+3{{x}^{3}}-7{{x}^{2}}-12x-4}}{{{{x}^{3}}-8}}\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2{{x}^{4}}+3{{x}^{3}}-7{{x}^{2}}-12x-4}}{{{{x}^{3}}-8}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}\left( {2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}} \right)}}{{{{x}^{3}}\left( {1-\frac{8}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{x\left( {2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}} \right)}}{{\left( {1-\frac{8}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,x\cdot \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}}}{{1-\frac{8}{{{{x}^{3}}}}}}\ $

$ \displaystyle =\infty $


8.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+3x+2}}{{{{x}^{3}}+32{{x}^{2}}-5x-20}}\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+3x+2}}{{{{x}^{3}}+32{{x}^{2}}-5x-20}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}\left( {1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}} \right)}}{{{{x}^{3}}\left( {1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\left( {1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}} \right)}}{{x\left( {1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{1}{x}\cdot \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}}}{{1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}}}\ $

$ \displaystyle =0$


9.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\left[ {\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\times \frac{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}} \right]\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+2x+1-({{x}^{2}}+1)}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{\sqrt{{{{x}^{2}}\left( {1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}} \right)}}+\sqrt{{{{x}^{2}}\left( {1+\frac{1}{{{{x}^{2}}}}} \right)}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{x\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+x\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{x\left( {\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}} \right)}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{2}{{\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}}}$

$ \displaystyle =\frac{2}{{\sqrt{{1+0+0}}+\sqrt{{1+0}}}}$

$ \displaystyle =\frac{2}{{\sqrt{1}+\sqrt{1}}}$

$ \displaystyle =1$


10.      $ \displaystyle \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+3+...+n}}{{{{n}^{2}}}}$

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Since $ \displaystyle 1 + 2 + 3 + … + n$ is an arithmetic series with the first term $ \displaystyle 1$ and the common difference $ \displaystyle 1$ having $ \displaystyle n$ terms.

$ \displaystyle \therefore 1+2+3+\ldots +n=\frac{n}{2}(1+n)$

$ \displaystyle \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+3+...+n}}{{{{n}^{2}}}}$

$ \displaystyle =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{{n+{{n}^{2}}}}{2}}}{{{{n}^{2}}}}$$ \displaystyle =\frac{1}{2}\cdot \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{n+{{n}^{2}}}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\cdot \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{n+{{n}^{2}}}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\underset{{n\to \infty }}{\mathop{{\cdot \lim }}}\,\frac{{{{n}^{2}}\left( {\frac{1}{n}+1} \right)}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\underset{{n\to \infty }}{\mathop{{\cdot \lim }}}\,\frac{{\frac{1}{n}+1}}{1}$

$ \displaystyle =\frac{1}{2}\left( 1 \right)$

$ \displaystyle =\frac{1}{2}$


11.      $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+4+...+{{2}^{{n-1}}}}}{{{{2}^{n}}+1}}$

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Since $ \displaystyle 1 + 2 + 4 + … + 2^{n – 1}$ is a geometric series with the first term $ \displaystyle 1$ and the common ratio $ \displaystyle 2$ having $ \displaystyle n$ terms.

$ \displaystyle \therefore 1+2+4+\ldots +{{2}^{{n-1}}}=\frac{{1({{2}^{n}}-1)}}{{2-1}}={{2}^{n}}-1$

$ \displaystyle \ \ \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+4+...+{{2}^{{n-1}}}}}{{{{2}^{n}}+1}}\ \ \ \ $

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{2}^{n}}-1}}{{{{2}^{n}}+1}}$

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{2}^{n}}(1-\frac{1}{{{{2}^{n}}}})}}{{{{2}^{n}}(1+\frac{1}{{{{2}^{n}}}})}}$

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1-\frac{1}{{{{2}^{n}}}}}}{{1+\frac{1}{{{{2}^{n}}}}}}$

$ \displaystyle \ \ \ =\frac{{1-0}}{{1+0}}$

$ \displaystyle \ \ \ =1$


12.      $ \displaystyle \ \underset{{t\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{t}+{{t}^{2}}}}{{2t-{{t}^{2}}}}$

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Let $ \displaystyle x=\sqrt{t}$ then $ \displaystyle t = x^2$

As $ \displaystyle t\to \infty, t\to \infty $

$ \displaystyle \therefore \ \ \ \ \ \ \underset{{t\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{t}+{{t}^{2}}}}{{2t-{{t}^{2}}}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{x+{{x}^{4}}}}{{2x-{{x}^{4}}}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}(\frac{1}{{{{x}^{3}}}}+1)}}{{{{x}^{4}}(\frac{2}{{{{x}^{3}}}}-1)}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+1}}{{\frac{2}{{{{x}^{3}}}}-1}}$

$ \displaystyle \ \ \ =\frac{{0+1}}{{0-1}}$

$ \displaystyle \ \ \ =-1$


13.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-{{4}^{{-x}}}}}{{{{4}^{x}}+{{4}^{{-x}}}}}$

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$ \displaystyle \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-{{4}^{{-x}}}}}{{{{4}^{x}}+{{4}^{{-x}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-\displaystyle \frac{1}{{{{4}^{x}}}}}}{{{{4}^{x}}+\displaystyle \frac{1}{{{{4}^{x}}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\left( {{{4}^{x}}-\displaystyle \frac{1}{{{{4}^{x}}}}} \right)\times \displaystyle \frac{1}{{{{4}^{x}}}}}}{{\left( {{{4}^{x}}+\displaystyle \frac{1}{{{{4}^{x}}}}} \right)\times \displaystyle \frac{1}{{{{4}^{x}}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{1-\displaystyle \frac{1}{{{{4}^{{2x}}}}}}}{{1+\displaystyle \frac{1}{{{{4}^{{2x}}}}}}}$

$ \displaystyle \begin{array}{l}=\ \ \displaystyle \frac{{1-0}}{{1+0}}\\\\=1\end{array}$