a limit without using l'hospital rule: limit, x->3, of (log_3(x) - 1)/(x - 3)

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[math]\lim_{x\to\infty}\frac{\log_3(x)-1}{x-3}\\\\\text{can someone give me a hint, how to calculate the limit without using l'Hospital rule - using this method it's easy, but without it...}[/math][math]\text{I don't need a solution, just a hint that will get me on the right track}[/math]

 

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[math]\text{there should be}\,\,\lim_{x\to3}...[/math]

 

[BigBeachBanana]

[math]\lim_{x\to 3}\frac{\log_3(x)-1}{x-3}\\\\\text{can someone give me a hint, how to calculate the limit without using l'Hospital rule - using this method it's easy, but without it...}[/math][math]\text{I don't need a solution, just a hint that will get me on the right track}[/math]

[math]\log_3x = \dfrac{\ln x}{\ln 3}[/math]
Try Taylor expansion for [imath]\ln(x)[/imath]

 

[topsquark]

[math]\lim_{x\to\infty}\frac{\log_3(x)-1}{x-3}\\\\\text{can someone give me a hint, how to calculate the limit without using l'Hospital rule - using this method it's easy, but without it...}[/math][math]\text{I don't need a solution, just a hint that will get me on the right track}[/math]

As there is no way to expand [imath]\log_b(x+h)[/imath] without a Taylor expansion, that's the only way you are going to be able to do this "by hand."

-Dan

 

[blamocur]

[math]\log_3x = \dfrac{\ln x}{\ln 3}[/math]
Try Taylor expansion for [imath]\ln(x)[/imath]

Or replacing [imath]x = 3^{y+1}[/imath] with [imath]y\rightarrow 0[/imath] and applying Taylor expansion there.

 

[Dr.Peterson]

[math]\lim_{x\to\infty}\frac{\log_3(x)-1}{x-3}\\\\\text{can someone give me a hint, how to calculate the limit without using l'Hospital rule - using this method it's easy, but without it...}[/math][math]\text{I don't need a solution, just a hint that will get me on the right track}[/math]

It depends on what limits you have learned, that you can use for this. You've only said (or implied) what you are not allowed to use.

If you know that [imath]\lim_{u\to0}\frac{\ln(u+1)}{u}=1[/imath], you can use that. Have you?

Or you could derive this fact from the derivative of the natural log.

 

[blamocur]

Try Taylor expansion for ln⁡(x)\ln(x)ln(x)

But Taylor expansion relies on derivatives, just like l'Hospital rule

 

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[math]\text{Calculate a limit without using l'Hopital's rule, so we can use derivatives.}\\\\\text{let}\,\,x=3^{y+1}\,\,\text{since}\,\,x\to3\,\,\text{then}\,\,y+1=1\,\,\Longleftrightarrow y=0\,\,\text{so we got}\\\\\lim_{x\to3}\frac{\log_3(x)-1}{x-3}=\lim_{y\to0}\frac{y+1-1}{3^{y+1}-3}=\lim_{y\to0}\frac{y}{3(3^y-1)}=\frac{1}{3}\lim_{y\to0}\left[\left(\frac{3^y-1}{y}\right)^{-1}\right]\stackrel{\red{**}}{=}\frac{1}{3\ln(3)}\\\\\red{**}\,\,\text{using the limit}\,\,\lim_{x\to0}\frac{a^x-1}{x}=\ln(a)\\\\\text{I think that I'm right. Thank to all of you guys, for your time and help!}[/math]