# A Horrible looking limit

I have the following limit question:

$$\lim_{x \rightarrow 1 }\frac {({\rm log} (1+x)-{\rm log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\rm sin}\space \pi x}$$ This has the form $\frac {0}{0}$ and hence I can apply L'Hospital's Rule. But, this is too large and that way it can go terribly wrong. Is there any clever way to do this?
I found it in a MCQ test and it has the following four alternatives:
A. $\frac{9}{\pi}{\rm log}\space \frac{4}{e}$
B. 1
C. $\frac{3}{\pi}{\rm log} \space \frac{2}{e}$
D. $\frac{1}{\pi}$.

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You can write $\frac {({\rm log} (1+x)-{\rm log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\rm sin}\space \pi x}=\frac{\log(1+x)-\log(2)}{x-1}\frac{3 4^{x-1}-3x}{x-1}\frac{x-1}{(7+x)^{1/3}-(1+3x)^{1/2}}\frac{x-1}{sin(\pi x)}=\frac{f(x)-f(1)}{x-1}\frac{g(x)-g(1)}{x-1}\frac{x-1}{h(x)-h(1)}\frac{x-1}{k(x)-k(1)}$

with $f(x)=\log(1+x)$, $g(x)=3 4^{x-1}-3x$, $h(x)=(7+x)^{1/3}-(1+3x)^{1/2}$ and $k(x)=sin(\pi x)$.

So $\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=f'(1)=\frac{1}{2}$, $\lim_{x\to1}\frac{g(x)-g(1)}{x-1}=g'(1)=3(\ln(4)-1)$, $\lim_{x\to1}\frac{h(x)-h(1)}{x-1}=h'(1)=\frac{1}{3}7^{-2/3}-\frac{3}{4}$ and $\lim_{x\to1}\frac{k(x)-k(1)}{x-1}=k'(1)=\pi\cos(\pi)=-\pi$.

Since $h'(1)\neq 0$ and $k'(1)\neq 0$, Limit Laws give you that the limit you're looking for is $\frac{f'(1)g'(1)}{h'(1)k'(1)}$.

PS: I don't know if $\log$ stands for logarithm in base $10$ or $e$ so I left as it is.

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Hint Apply L'H to both $$\lim_{x \rightarrow 1 }\frac {{\rm log} (1+x)-{\rm log}\space 2}{{\rm sin}\space \pi x}$$ and $$\lim_{x \rightarrow 1 }\frac {4^{x-1}-1}{(7+x)^{1/3}-(1+3x)^{1/2}}$$

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Put $x=y+1$ and take the limit $y\to 0$, so that we can use some standard limits and simplify the expression. The given limit will be:

\begin{align*} & \lim_{y\to 0} - \frac{\displaystyle \log\left(1+\frac{y}{2}\right)\Big(3\, \left(4^y-1-y\right)\Big)}{\left(\left(8+y\right)^{1/3}-\left(4+3\, y\right)^{1/2}\right)\, \sin\left(\pi\, y\right)}\tag 1\\ \end{align*} Divide the numerator and denominator by $y^2$ and we will take four separate limits:

\begin{align*} \lim_{y\to 0} \frac{\displaystyle \log\left(1+\frac{y}{2}\right)}{\displaystyle \frac{y}{2}\cdot 2} &= \frac{1}{2}\\\\ \lim_{y\to 0} \frac{4^y-1}{y}-1 &= \log{4}-1\\\\ \lim_{y\to 0} \frac{\sin\left(\pi\, y\right)}{\pi\, y}\cdot \pi &= \pi \end{align*} and for the remaining one we can use L'Hôpital's rule: \begin{align*} \lim_{y\to 0} \frac{\left(8+y\right)^{1/3}-\left(4+3\, y\right)^{1/2}}{y} &= \lim_{y\to 0} \frac{1}{3}\left(8+y\right)^{-2/3}-\frac{3}{2}\left(4+3\, y\right)^{-1/2}\\ &= \frac{1}{3\cdot 4}-\frac{3}{4} = -\frac{2}{3}\\ \end{align*} Combining all these in $(1)$, we see that

\begin{align*} \lim_{x \to 1 }\frac {({\log} (1+x)-{\log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\sin}\space \pi x} = \frac{9}{4\, \pi}\log\left(\frac{4}{e}\right)\approx 0.276662956773403 \end{align*}

which means none of the options are correct.

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