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The question is

$$\lim_{x\to0}\frac{ 5^x-3^x}{x}$$

Unfortunately we cannot use the L'Hôpital's rule to solve this limit so I'm not sure how to proceed with the question.

I'm sure there is an elegant way to manipulate the algebra to resolve the indeterminate form, however I just cant find it. Any help would be much appreciated thanks again.

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  • $\begingroup$ You can try writing using the Taylor series expansion on $5^{x}$ and $3^{x}$. $\endgroup$ – user262291 Oct 10 '16 at 1:50
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Remember the fundamental limits $\lim_{x \rightarrow 0} \frac{a^x -1}{x} = \ln(a)$.

Then write $\frac{5^x - 3^x}{x} = \frac{5^x - 1}{x} - \frac{3^x -1}{x}$.

So you have that $\lim_{x \rightarrow 0}\frac{5^x - 1}{x} - \frac{3^x -1}{x} = \lim_{x \rightarrow 0}\frac{5^x - 1}{x} - \lim_{x \rightarrow 0}\frac{3^x - 1}{x} = \ln(5) - \ln(3) = \ln(\frac{5}{3}) $

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Let us consider the most general case of $$\lim_{x\to0}\frac{ a^x-b^x}{x}$$ $$\frac{ a^x-b^x}{x}=\frac{ e^{x \log(a)}- e^{x \log(b)}}{x} $$ For any value of $t$ we have $$e^t=\sum_{i=0}^\infty \frac{t^n}{n!}$$ So, replace $t$ by $x\log(a)$ and later by $x\log(b)$. This makes $$a^x-b^x=\left(1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+O\left(x^3\right) \right)-\left(1+x \log (b)+\frac{1}{2} x^2 \log ^2(b)+O\left(x^3\right) \right)$$ $$a^x-b^x=x (\log (a)-\log (b))+\frac{1}{2} x^2 \left(\log ^2(a)-\log ^2(b)\right)+O\left(x^3\right)$$ $$\frac{ a^x-b^x}{x}=(\log (a)-\log (b))+\frac{1}{2} x \left(\log ^2(a)-\log ^2(b)\right)+O\left(x^2\right)$$ $$\frac{ a^x-b^x}{x}=\log \left(\frac{a}{b}\right)\left(1+\log(\sqrt{ab})x\right)+O\left(x^2\right)$$ which shows the limit and also how it is approached when $x\to 0$.

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