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I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.

$$\lim_{x\to\infty}x\log\left(\frac{x+7}{x+2}\right)$$

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  • $\begingroup$ Are we allowed to use L'Hospital Rule? $\endgroup$ – Sujit Bhattacharyya Dec 1 '18 at 12:48
  • $\begingroup$ I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way. $\endgroup$ – Michael Dec 1 '18 at 12:50
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    $\begingroup$ Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier. $\endgroup$ – Michael Dec 1 '18 at 13:02
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Use equivalents: we know that, near $0$, $\ln(1+u)\sim u$, so $$\ln\biggl(\frac{x+7}{x+2}\biggr)=\ln\biggl(1+\frac{5}{x+2}\biggr)\sim_{x\to +\infty}\frac{5}{x+2},$$ therefore $$x\ln\biggl(\frac{x+7}{x+2}\biggr)\sim_{x\to +\infty}\frac{5x}{x+2}\to 5.$$

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  • $\begingroup$ It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1). $\endgroup$ – gimusi Dec 1 '18 at 13:47
  • $\begingroup$ Simple example since $\cos x\sim 1$ we could erroneously conclude that $(1-\cos x)/x^2\sim 0$ which is wrong of course. $\endgroup$ – gimusi Dec 1 '18 at 13:49
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Since $$\log(1+t) \sim t$$ as $t \to 0$, we have that

$$\lim_{x \to \infty} x \log \left(1 + \frac 5{x+2}\right) = \lim_{x \to \infty} x\cdot \frac 5{x+2} = 5$$

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As an alternative way, you can also note that

$$x\ln\bigg(\frac{x+7}{x+2}\bigg) = \ln\bigg(\frac{x+7}{x+2}\bigg)^x = \ln\bigg(1+\frac{5}{x+2}\bigg)^x$$

$$= \ln\Biggl[\bigg(1+\frac{5}{x+2}\bigg)^{x+2}\Biggl]^{\frac{x}{x+2}}$$

and

$$\lim_{n\to \infty} \bigg(1+\frac{x}{n}\bigg)^n = e^x$$

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First note that, $$\lim_{x\to 0}\frac{\log(1+mx)}{x}=m$$ for any real $m$.

Putting, $x=1/y$ we get, $x\to\infty\implies y\to 0$ and the limit becomes, $$\lim_{x\to\infty}x\log\left(\frac{x+7}{x+2}\right)=\lim_{y\to0}\frac{1}{y}\log\left(\frac{1+7y}{1+2y}\right)=\lim_{y\to0}\frac{1}{y}\{\log(1+7y)-\log(1+2y)\}=7-2=5$$

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    $\begingroup$ It's trivially true of $m=0$ as well, because of $\frac{0}{x} \to 0$. $\endgroup$ – Botond Dec 1 '18 at 13:09
  • $\begingroup$ Thanks for pointing that. I edited. $\endgroup$ – Sujit Bhattacharyya Dec 1 '18 at 13:18
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$\lim_{x \rightarrow \infty} \log(\dfrac{(1+7/x)^x}{(1+2/x)^x})=$

$\log \dfrac{\lim_{x \rightarrow \infty}(1+7/x)^x}{\lim_{x \rightarrow \infty}(1+2/x)^x}=$

$\log \left(\dfrac{e^7}{e^2}\right)= 5.$

Used :

$\lim_{x \rightarrow \infty} (1+a/x)^x=e^a$, $a$ real.

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HINT

$$x\log\left(\frac{x+7}{x+2}\right)=\log \left(1 + \frac 5{x+2}\right)^x$$

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$\displaystyle \lim_{x \to \infty} x \log\biggr(\frac{x+7}{x+2}\biggl)$ is of the indeterminate form $0 \cdot \infty$. We can rewrite the expression to be of the form $\dfrac{\infty}{\frac{1}{0}}$ to be able to apply L'Hopital's Rule.

$$x \log \biggr(\frac{x+7}{x+2}\biggl)= \dfrac{\log (\frac{x+7}{x+2})}{\frac{1}{x}} \implies \displaystyle \lim_{x \to \infty} \dfrac{\log (\frac{x+7}{x+2})}{\frac{1}{x}} = \displaystyle \lim_{x \to \infty} \dfrac{\frac{x+2}{x+7}}{\frac{-1}{x^2}}\cdot \frac{d}{dx}\biggr(\frac{x+7}{x+2}\biggl)$$

Thus the limit translates to the new equivalent simplified expression i.e. $\displaystyle \lim_{x \to \infty} \dfrac{5x^2}{(x+7)(x+2)}$.

Therefore, $\displaystyle \lim_{x \to \infty} x \log \biggr(\frac{x+7}{x+2}\biggl)=5$. Cheers!

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