1

My Teacher has told to use this as a standard result so when question is in this I directly apply this result as in this

$$\lim\limits_{x\rightarrow 0}\dfrac{\log\left( 5+x\right) }{x}-\dfrac{\log\left( 5-x\right) }{x}=2/5$$ but here

$$\lim\limits_{x\rightarrow 0}\dfrac{\log\left( 1+2x\right) -2{\log\left( 1+x\right)} }{x^2}=0 $$

doesn't work because solution is give using expansion of $\log\left( 1+x\right)$ I am interested to know why we cant simply multiply and divide 2x to log(1+2x) to convert it in known form like log(1+2x) / 2x *2x yielding 2x So what is limitation of result when to use it and when to not?

Ha'Penny
  • 59
  • 2
    I'm completely at a loss for what you have done. – calc ll Jul 01 '22 at 03:40
  • Remember that the standard limit works everywhere and it means that whenever you see $\lim_{x\to 0}\frac{\log(1+x)}{x}$ you can replace it with $1$. It does not mean that you can replace $\frac{\log(1+x)}{x}$ with $1$ or $\log(1+x)$ by $x$. Mathematically "$\lim_{x\to a} f(x) $" is a very different beast compared to $f(x) $. – Paramanand Singh Jul 05 '22 at 06:56
  • See more details at https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Jul 05 '22 at 06:59

3 Answers3

3

In the first case, you have the limit $$\lim_{x\to 0}\frac{\ln(5+x)-\ln(5-x)}{x}.$$ Since $\lim\limits_{x\to 0}(\ln(5+x)-\ln(5-x))=0$ and $\lim\limits_{x\to 0}(x)=0$, L'Hopital's rule says that the above limit is equal to $$\lim_{x\to 0}\left(\frac{1}{x+5}+\frac{1}{5-x}\right)=\frac25.$$

(I wouldn't call this a direct application of $\lim\limits_{x\to 0}\frac{\ln(1+x)}{x}=1$, but it's possible there's another solution using that fact.)

For the second problem, we again have the so-called indeterminate form $\frac00$, so applying L'Hospital's rule gives $$\lim_{x\to 0}\frac{\frac{2}{1+2x}-\frac{2}{1+x}}{2x}=\lim_{x\to 0}\frac{\frac{-2x}{(1+2x)(1+x)}}{2x}=\lim_{x\to 0}\frac{-1}{(1+2x)(1+x)}=-1.$$

pancini
  • 19,216
  • i'm rather interested in application of $\lim\limits_{x\to 0}\frac{\ln(1+x)}{x}=1$ – Ha'Penny Jul 01 '22 at 05:09
  • @Ha'Penny it isn't clear to me how you would directly use that fact here, but maybe it's possible. It would probably not be a very clean or easy solution though. – pancini Jul 01 '22 at 05:14
  • @Sung jin-woo 's post would be the way to do this. There's a small technicality to deal with when you replace $x$ with $2x/(5-x)$, but if this is a high school level class you probably don't need to worry about it. (Sorry if this doesn't render correctly; i'm on mobile) – pancini Jul 01 '22 at 10:14
  • @pancini Out of interest , would you please give some idea as to the technicality involved in the substitution – Sung Jin-Woo Jul 01 '22 at 14:57
2

Though L'Hospitals rule as used by @pancini is a better alternative to your problem, I would just show that the aforementioned 'result' can be used on both the problems.

First problem: Rewrite as $$\lim_{x\to 0} \frac{\ln\frac{5+x}{5-x}}{x}$$

The numerator is $$\ln(1+ \frac{2x}{5-x})$$

Hence the limit becomes $$\lim_{x \to 0} \frac{\ln(1+ \frac{2x}{5-x})}{(\frac{2x}{5-x})} × \frac{2}{5-x} $$

Here I use your result ,but take the expression$$\frac{2x}{5-x}$$which also tends to zero,hence your "result" is applicable modifying it as,$$ \lim \frac{2}{5-x} = \frac{2}{5} $$

In the second problem you proceed exactly as above , except at a particular step you substitute $x^2=t$ Then the standard result will show up. I will leave that calculation to yourself.

Sung Jin-Woo
  • 562
  • undoubtedly this one of many methods. But I am interested to know why we cant simply multiply and divide 2x to log(1+2x) to convert it in known form like log(1+2x) / 2x *2x yielding 2x – Ha'Penny Jul 01 '22 at 05:13
  • We can do so. In fact it is pretty much the same of what I have done in the above post ( see the line "Hence the limit becomes..." In the post. We can't just directly do it in your second problem because of the second term in the numerator, ( which is not "compatible" with $2x$. ) Hence it would result in an indeterminate form – Sung Jin-Woo Jul 01 '22 at 05:20
  • In your problem , try putting the two logs together and then seperate the expression in a $\ln 1+(0)$ form and substitute $x^2=t$ and then proceed in a similar fashion as in my post. If I still unclear, I'll edit my answer to accomodate the second solution – Sung Jin-Woo Jul 01 '22 at 05:26
  • @Sungjin-woo Write $f(x)=\frac{\ln(x+1)}{x}$ and $g(x)=\frac{2x}{5-x}$. Then we know $\lim\limits_{x\to 0}f(x)=1$ and $\lim\limits_{x\to 0}g(x)=0$. We want to conclude $\lim\limits_{x\to 0}f(g(x))=\lim\limits_{x\to 0}f(x)$ since in either case the thing inside $f$ is going to $0$. This is not true in general since $g(x)$ might actually be equal to $0$ at too many points. I can say more in chat if you want. – pancini Jul 02 '22 at 04:00
  • @Sungjin-woo (Also I meant to reply to your comment on my answer asking for clarification.) – pancini Jul 02 '22 at 08:00
  • @pancini thank you! You mean like those wildly oscillating function which tend to zero right – Sung Jin-Woo Jul 02 '22 at 14:08
  • @Sungjin-woo Right; you can even have the inside function be continuous. E.g. let $f(x)=x\sin(\pi/x)$ and $g(x)=1$ for all $x\neq 0$, and set $f(0)=0$ and $g(0)=0$. Then for all $n\in\Bbb N$, $g\circ f(1/n)=0$ but $g\circ f(\sqrt 2/n)=1$. – pancini Jul 05 '22 at 05:51
0

The proposed result is not valid.

Having said that, you may be also interested in the power series method: \begin{align*} \ln(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \ldots \end{align*} whenever $|x| < 1$. If we restrict the values of $x$ to $(-1/2,1/2)$, we may claim as well that

\begin{align*} \ln(1 + 2x) = 2x - 2x^{2} + \frac{8x^{3}}{3} - \ldots \end{align*}

Therefore the proposed limit is given by \begin{align*} \lim_{x\to 0}\frac{\ln(1 + 2x) - 2\ln(1 + x)}{x^{2}} & = \lim_{x\to 0}\frac{2x - 2x^{2} - 2x + x^{2} + o(x^{2})}{x^{2}} = \lim_{x\to 0}\frac{-x^{2} + o(x^{2})}{x^{2}} = -1 \end{align*}

Hopefully this helps!

Átila Correia
  • 15,530