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How to find this limit without L'Hopital

$\lim_{x \to 0}\Bigl(\dfrac{3^x-5^x}{x}\Bigr)$ ??

Challenger5
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Klaus82
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5 Answers5

7

HINT:

Note that we have

$$\lim_{x\to 0}\frac{3^x-5^x}{x}=\lim_{h\to 0}\left(\frac{3^h-3^0}{h}-\frac{5^h-5^0}{h}\right)$$

Use the definition of the derivative.

Mark Viola
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  • If resourcing to derivatives is allowed, why not L'Hospital? It seems too restrictive, though perhaps some teacher may want the students to practice this way – DonAntonio Feb 22 '18 at 21:26
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    @DonAntonio Differentiation is taught before introducing L'Hospital's Rule. I've posted a second answer that doesn't rely at all of derivatives. – Mark Viola Feb 22 '18 at 21:31
  • @MarkViola I agree with you, best way when possible to use it. – user Feb 22 '18 at 21:43
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    @DonAntonio The given limit is essentially equivalent to computing $\lim_{x\to0}(a^x-1)/x$, which is just a derivative. A good teacher should emphasize the fact that $\lim_{x\to0}(e^x-1)/x$ is the derivative at zero of the exponential function; likewise for $\lim_{x\to0}(\sin x)/x$. – egreg Feb 22 '18 at 21:58
  • Small critique: You don't know the limit exists until you prove it does. Why write $\lim_{x\to 0}$ so early? You certainly don't know you can write the limit, which you don't know exists yet, as the difference of two other limits, which you haven't shown exist yet. I would get rid of the limit signs at the beginning and simply write$$\frac{3^x-5^x}{x} = \frac{3^x-3^0}{x} - \frac{5^x-5^0}{x}.$$Now you're good to go with your explanation. – zhw. Feb 22 '18 at 22:42
  • @Zhw. Yes, I know. But the word HINT preceeds the development. So, I took the liberty of proceeding carelessly. – Mark Viola Feb 22 '18 at 22:52
  • OK, good point. – zhw. Feb 22 '18 at 23:13
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By standard limit (and derivative definition)

$$\frac{a^x-1}{x}\to \log a$$

we have

$$\dfrac{3^x-5^x}{x}=\dfrac{3^x-1}{x}-\dfrac{5^x-1}{x}\to\log 3-\log 5=\log\frac 35$$

user
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You can use the expansion of $a^x$ to expand $3^x$ and $5^x$. Then proceed further. Hint $a^x = Exp(xlna)$.

taniya kapoor
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Same idea than in other answer: $$ \lim_{x\to 0}\frac{3^x - 5^x}{x} = -\lim_{x\to 0}\frac{(5/3)^x - 1}{x}3^x = -\lim_{x\to 0}\frac{(5/3)^x - 1}{x}\lim_{x\to 0}3^x $$ and apply the definition of derivative.

Martín-Blas Pérez Pinilla
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I remember back when I took calculus that we were given the definition $$\ln x=\int_1^x \frac{dt}t$$ And then $\exp(x)=e^x$ was the inverse function to that and we were supposed to prove everything from that. In this case from the laws of logarithms and exponents, $$\frac{3^x-5^x}x=-3^x\frac{\left(\frac53\right)^x-1}x=-3^x\frac{\exp\left(x\ln\frac53\right)-1}x$$ Then if $y=\exp\left(x\ln\frac53\right)$, then $$\ln y=x\ln\frac53=\int_1^y\frac{dt}t$$ if $x>0$, then $1/y\le1/t\le1$ over the interval of integration, so $$1-\frac1y=\frac1y(y-1)\le x\ln\frac53\le y-1$$ So $$x\ln\frac53+1\le y\le\frac1{1-x\ln\frac53}=\frac{x\ln\frac53}{1-x\ln\frac53}+1$$ So that $$\ln\frac53\le\frac{y-1}x=\frac{\exp\left(x\ln\frac53\right)-1}x\le\frac{\ln\frac53}{1-x\ln\frac53}$$ So it follows from the squeeze theorem that $$\lim_{x\rightarrow0^+}\frac{\exp\left(x\ln\frac53\right)-1}x=\ln\frac53$$ If $x<0$, then $$x\ln\frac53=\int_1^y\frac{dt}t=-\int_y^1\frac{dt}t$$ And $$1-y\le\int_y^1\frac{dt}t=-x\ln\frac53\le\frac1y(1-y)$$ Which works out to $$\ln\frac53\ge\frac{y-1}x=\frac{\exp\left(x\ln\frac53\right)-1}x\ge\frac{\ln\frac53}{1-x\ln\frac53}$$ So that $$\lim_{x\rightarrow0^-}\frac{\exp\left(x\ln\frac53\right)-1}x=\ln\frac53$$ The same approach shows also that $$\lim_{x\rightarrow0}\exp\left(x\ln3\right)=1$$ So $$\lim_{x\rightarrow0}\frac{3^x-5^x}x=\lim_{x\rightarrow0}-\exp\left(x\ln3\right)\frac{\exp\left(x\ln\frac53\right)-1}x=-(1)\left(\ln\frac53\right)=-\ln\frac53$$

user5713492
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