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I am trying to calculate the following limits

$$ \lim_{x\to\infty}(2x+1) \ln \left(\frac{x-3}{x+2}\right) $$ and $$ \lim_{x\to1}\frac{x}{3x-3}\ln(7-6x) $$

I can't use l'Hôpital's rule, so I am not sure how to solve it.

soupless
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Daniel
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  • I don't know why the aberrance to using a nice result as L'Hospital. Anyway, for the first limit, notice $(2x+1)\log\big(\frac{x-3}{x+2}\big)=\log\Big(1-\frac{-5}{x+2}\Big)^{2x+1}$. Recall that $(1+a/x)^x\xrightarrow{x\rightarrow\infty}e^a$. Te second can be done similarly, modulo a change of variables, or by using Taylor expansions around $x=1$. – Mittens Apr 19 '21 at 17:01
  • @OliverDiaz Thank you. I managed to calculate the first limit. Can you explain about the second one? – Daniel Apr 19 '21 at 17:14
  • use $\ln(1+x)/x\to 1$ in $0$. hint $7-6x=1+6(1-x))$. – zwim Apr 19 '21 at 17:17
  • The function in question has a singularity at $x=1$ (both numerator and numerator vanish at $1$. Use Taylor series of $\log(7-6x)=\log(7-6(x-1+1))=\log(1-6(x-1))$ to see if there are cancellations. You need to do the work to really lear:

    Hint: $\log(1-x)=-\sum_{n\geq1}\frac{1}{n}(x-1)^n$

     – Mittens Apr 19 '21 at 17:19
  • @OliverDiaz I don't think it's that fair to use the Taylor series when the L'Hospital is not allowed. – VIVID Apr 19 '21 at 17:24
  • Then either you take swim as an act of Fatih or use something related, $\lim_n n^p/(1+a)^n=0$ $(a>0)$ which is studies typically in Calculus classes. transform $x=(1+a)^n$ and see what happens. – Mittens Apr 19 '21 at 17:25

2 Answers2

1

By continuity of the logarithm,

$$ \lim_{x\to\infty} \ln \left(\frac{x-3}{x+2}\right)^{2x+1} = \lim_{x\to\infty} \ln \left(1-\frac5{x+2}\right)^{2(x+2)-3}\\ =\ln((e^{-5})^2)-3\lim_{x\to\infty}\ln \left(1-\frac5{x+2}\right).$$

-3

$$\lim_{x\rightarrow \infty }\left ( 2x+1 \right )\ln \left ( \frac{x-3}{x+2} \right )=\lim_{x\rightarrow \infty }\frac{\frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( \frac{x-3}{x+2} \right )}{\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{\left ( 2x+1 \right )}}=-\lim_{x\rightarrow \infty }\frac{5\left ( 2x+1 \right )^2}{2\left ( x-3 \right )\left ( x+2 \right )}=-\frac{5}{2}\lim_{x\rightarrow \infty }\frac{\left ( 2x+1 \right )^2}{\left ( x-3 \right )\left ( x+2 \right )}=-\frac{5}{2}\cdot 4\lim_{x\rightarrow \infty }\frac{2x+1}{2x-1}=-\frac{5}{2}\cdot 4=-10$$
$$\lim_{x\rightarrow 1 }\frac{x}{3x-3}\ln \left ( 7-6x \right )=\lim_{x\rightarrow 1 }\frac{\ln \left ( 7-6x \right )}{3x-3}=\lim_{x\rightarrow 1 }\frac{\frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( 7-6x \right )}{\frac{\mathrm{d} }{\mathrm{d} x}3x-3}=\lim_{x\rightarrow 1 }\frac{2}{6x-7}=-2$$

Dmitry
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