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# limit

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limit as x approaches 0 for (3^x-10^x)/x

Guest Feb 15, 2017
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#1
+10613
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Not sure       -1.2039????      (L'Hopital rule)

ElectricPavlov  Feb 15, 2017
#2
+89749
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limit as x approaches 0 for (3^x-10^x)/x

The numerator and the denominator both tend to 0 so I am going to try l'hopital's rule

mmm

$$let\;\;\\ t=3^x\\ ln_3t=x\\ x=\frac{lnt}{ln3}\\ \frac{dx}{dt}=\frac{1}{tln3}\\ \frac{dt}{dx}=tln3\\ \frac{dt}{dx}=3^xln3\\$$

Using the same method

$$\frac{d}{dx}(10^x)=10^xln10$$

$$\displaystyle\lim_{x\rightarrow 0}\;\;\frac{3^x-10^x}{x}\\ =\displaystyle\lim_{x\rightarrow 0}\;\;\frac{3^xln3-10^xln10}{1}\\ =\displaystyle\lim_{x\rightarrow 0}\;\;3^xln3-10^xln10\\ =ln3-ln10\\~\\ =ln(0.3)\\~\\ \approx -1.204$$

Here is the graph

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Melody  Feb 15, 2017
#3
+5

Find the following limit:
lim_(x->0) (3^x - 10^x)/x

Applying l'HÃ´pital's rule, we get that
lim_(x->0) (3^x - 10^x)/x | = | lim_(x->0) ( d/( dx)(3^x - 10^x))/(( dx)/( dx))
| = | lim_(x->0) (3^x log(3) - 10^x log(10))/1
| = | lim_(x->0) (3^x log(3) - 10^x log(10))
lim_(x->0) (log(3) 3^x - log(10) 10^x)

lim_(x->0) (3^x log(3) - 10^x log(10)) = 3^0 log(3) - 10^0 log(10) = -log(10/3):