2
$\begingroup$

Use the definition of derivative and find the following limit:

$\lim_{x\to0} \dfrac{\ln(2x+1)-\ln(1-3x)}{x}$

I do not understand what this question is asking me to do.

What does it mean to get the limit at 0 and how does that relate to the derivative using this example?

Are not the limit and the derivative at 0 going to be different?

I am really confused as to how I need to approach this question, do I take the derivative of the limit at 0?

I am probably misinterpreting this question altogether, please help me clarify? Thank you.

$\endgroup$
  • $\begingroup$ Please learn to use MathJax, as stated on the ask-a-question page. $\endgroup$ – user21820 Oct 31 '18 at 7:51
  • $\begingroup$ Thank you, I will next time! $\endgroup$ – Josh Teal Oct 31 '18 at 14:45
4
$\begingroup$

Notice that $ \ln (2x + 1 ) - \ln (1-3x) = \ln \left( \frac{2x+1}{1-3x} \right )$. Let $f(x) = \ln \left( \frac{2x+1}{1-3x} \right )$ and $f(0) = \ln 1 = 0 $. Now, your limit reads as

\begin{align*} \lim_{x \to 0} \dfrac{\ln (2x + 1 ) - \ln (1-3x)}{x} &= \lim_{x \to 0} \frac{ \ln \left( \frac{2x+1}{1-3x} \right ) }{x} \\ &=\lim_{x \to 0} \frac{ f(x) - f(0) }{x-0} \\ &= f'(0) \end{align*}

Can you finish it??

$\endgroup$
  • $\begingroup$ For clarification - we are manipulating our expression to look like a definition of a derivative but still solving the limit at 0, right? Also, other than the purpose of being asked, why would we want to do this? $\endgroup$ – Josh Teal Oct 31 '18 at 5:56
  • $\begingroup$ Also, since you manipulative the expression to fit the def of a derivative does this mean the limit is the same as the derivative at x=0? $\endgroup$ – Josh Teal Oct 31 '18 at 6:11
4
$\begingroup$

Hint:

Use the following property, if $f$ is differentiable,

$$\lim_{h \to 0 } \frac{f(y+mh) - f(y-nh)}{(m+n)h}=f'(y)$$

Edit:

If $f$ is differentiable,

$$\lim_{h \to 0} \frac{f(y+h)-f(y)}{h}=f'(x) = \lim_{h \to 0}\frac{f(y)-f(y-h)}{h}$$

$$\lim_{h \to 0} \frac{f(y+mh)-f(y)}{mh}=f'(x) = \lim_{h \to 0}\frac{f(y)-f(y-nh)}{nh}$$

\begin{align}\lim_{h \to 0} \frac{f(y+mh) -f(y-nh)}{(m+n)h} &=\lim_{h \to 0} \frac{f(y+mh)-f(y)+f(y) -f(y-nh)}{(m+n)h}\\ &=\lim_{h \to 0} \frac{mh}{(m+n)h}\frac{f(y+mh)-f(y)}{mh}+\lim_{h \to 0} \frac{nh}{(m+n)h}\frac{f(y)-f(y-nh)}{nh}\\ &=\frac{m}{(m+n)}\lim_{h \to 0} \frac{f(y+mh)-f(y)}{mh}+\frac{n}{(m+n)}\lim_{h \to 0} \frac{f(y)-f(y-nh)}{nh}\\ &= \frac{m}{m+n}f'(y) + \frac{n}{m+n}f'(y)\\ &= f'(y)\end{align}

$\endgroup$
  • 1
    $\begingroup$ better to use the law of logs and use the usual definition of the derivative. See my answer below. (+1) $\endgroup$ – James Oct 31 '18 at 4:15
  • $\begingroup$ nice approach. =) $\endgroup$ – Siong Thye Goh Oct 31 '18 at 4:18
  • $\begingroup$ Can I ask you a question, since I know you are the Linear Programming/optimization guru around here, do you have any book recommendation or webpage with problems and solutions about LP? or just a problem book. $\endgroup$ – James Oct 31 '18 at 4:20
  • 1
    $\begingroup$ hmmm.... not really. this page has some recommendation. I browsed through the first few chapters of the book Introduction to Linear Optimization to prepare for my exam a few years ago. $\endgroup$ – Siong Thye Goh Oct 31 '18 at 4:31
  • $\begingroup$ I do not understand where that came from - are we suppose to know this property? $\endgroup$ – Josh Teal Oct 31 '18 at 5:57
2
$\begingroup$

Straightforward:

$F(x)=\ln (2x+1)- \ln (1-3x).$

$F(0)= 0.$

$\lim_{ x \rightarrow 0} \dfrac{F(x)-F(0)}{x-0}=F'(0)=$

$2 + 3= 5.$

Appended:

$F'(x) =$

$ (\log (2x+1))' - (\log (1-3x))'=$

$\dfrac{1}{2x+1} \cdot (2) - \dfrac{1}{1-3x} \cdot (-3)$.

$F'(0)= 2-(-3)=5.$

(Chain rule)

$\endgroup$
  • $\begingroup$ Can you please add more clarification as to how you found this limit(without using l'Hopital's rule)? Where did 2 +3 come from? $\endgroup$ – Josh Teal Oct 31 '18 at 14:47
  • $\begingroup$ Josh.Of course.I.put it in the answer.Give me a little time. $\endgroup$ – Peter Szilas Oct 31 '18 at 17:23
  • $\begingroup$ Josh. Used ( log x)' =1/x , and chain rule. First you differentiate with respect to the argument, 1st term is 1/(2x+1) and then multiply by d/dx (2x+1)=2.Your thoughts? $\endgroup$ – Peter Szilas Oct 31 '18 at 17:35
  • $\begingroup$ Oh, since you manipulated it to look like a derivative equation you can just use the derivative to calculate the limit at 0? But, saying this we can also simplify the expression and solve it like a limit at 0 right? $\endgroup$ – Josh Teal Oct 31 '18 at 21:14
  • $\begingroup$ Josh.Did not manipulate much, this is the definition of the derivative of F(x) at 0, since F(0)=0.This is one way.If you do not want to use the derivative , one can try other options to find the limit to zero, as I understood you wanted the derivative, which is straight forward here. Your thoughts? $\endgroup$ – Peter Szilas Oct 31 '18 at 21:27
0
$\begingroup$

We have

$$\lim_{x\to0} \frac{\ln(2x+1)-\ln(1-3x)}{x}=\lim_{x\to0} \frac{\ln(2x+1)-\ln 1}{x-0}-\lim_{x\to0} \frac{\ln(1-3x)-\ln 1}{x-0}$$$$=f’(0)-g’(0)=\left(\frac2{2x+1}\right)_{(x=0)}-\left(\frac{-3}{1-3x}\right)_{(x=0)}=2-(-3)=5$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.