Can anyone help me solve these two limits? I am learning limits on my own and I progress slowly.
$\lim\limits_{x\to \infty}\left ( \frac{1-3x}{1-2x} \right )^{\frac{-2x+1}{x}}$
$\lim\limits_{x\to0}\frac{x}{\tan( 2 x ) }$
Any help is welcome.
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It shall be $\tan$ or something. – PenasRaul Sep 12 '14 at 10:19
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@notorious, Can you please check the first limit, Should it not be $x\to0?$ – lab bhattacharjee Sep 12 '14 at 11:20
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Is it $x\to\infty$ or $x\to0$ in the first problem? If it is $0$, the limit can be gotten as in lab's answer. If it is $\infty$, the limit can be gotten as in my answer. – robjohn Sep 12 '14 at 11:20
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It is $x\to\infty$. – LearningMath Sep 12 '14 at 16:40
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1both ($x\to0$ and $x\to\infty$) are interesting problems, dealing with different aspects of limits of composite functions. – robjohn Sep 12 '14 at 16:45
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$$\frac{1-3x}{1-2x}-1=\frac{-x}{1-2x}$$
Use $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$
$$\lim_{x\to0}\frac x{\tan2x}=\lim_{x\to0}\cos2x\cdot\lim_{x\to0}\frac1{\dfrac{\sin2x}{2x}}\cdot\frac12$$
lab bhattacharjee
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@notorious, See https://proofwiki.org/wiki/Limit_of_Sine_of_X_over_X and https://proofwiki.org/wiki/Exponential_as_Limit_of_Sequence – lab bhattacharjee Sep 12 '14 at 10:37
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Dd you read the first problem correctly? The base tends to $\frac32$ and the exponent to $-2$. Ah, I see... $x$ is tending to $\infty$, not $0$. – robjohn Sep 12 '14 at 11:18
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Since $x^y$ is continuous near $x=\frac32$ and $y=-2$, we get $$ \begin{align} \lim_{x\to\infty}\left(\frac{1-3x}{1-2x}\right)^{\frac{-2x+1}{x}} &=\left(\frac32\right)^{-2}\\ &=\frac49 \end{align} $$ As shown in this answer, $\lim\limits_{x\to0}\dfrac{\tan(x)}{x}=1$, and since $\frac1x$ is continuous near $x=1$, we get $$ \begin{align} \lim_{x\to0}\frac{x}{\tan(2x)} &=\frac12\lim_{x\to0}\frac{2x}{\tan(2x)}\\ &=\frac12 \end{align} $$
