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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.

Community
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Josh Teal
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4 Answers4

5

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}

Siong Thye Goh
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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??

James
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  • 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? – Josh Teal Oct 31 '18 at 05:56
  • 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? – Josh Teal Oct 31 '18 at 06:11
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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)

Peter Szilas
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  • 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? – Josh Teal Oct 31 '18 at 14:47
  • Josh.Of course.I.put it in the answer.Give me a little time. – Peter Szilas Oct 31 '18 at 17:23
  • 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? – Peter Szilas Oct 31 '18 at 17:35
  • 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? – Josh Teal Oct 31 '18 at 21:14
  • 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? – Peter Szilas Oct 31 '18 at 21:27
  • No, I agree, thanks, I got it! – Josh Teal Nov 01 '18 at 22:43
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We have

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

user
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