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Numerade Educator



Problem 33 Medium Difficulty

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.

$ \displaystyle \lim_{x\to 0} \frac{x3^x}{3^x - 1} $


This limit has the form
$\frac{0}{0} \cdot \lim _{x \rightarrow 0} \frac{x^{3}}{3^{x}-1}=\lim _{x \rightarrow 0} \frac{x 3^{x} \ln 3+3^{x}}{3^{x} \ln 3}=\lim _{x \rightarrow 0} \frac{3^{x}(x \ln 3+1)}{3^{x} \ln 3}=\lim _{x \rightarrow 0} \frac{x \ln 3+1}{\ln 3}=\frac{1}{\ln 3}$


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Video Transcript

So we're gonna be using local tiles role in order to find the limit of the function. The function is X times three to the X. X Times 3 to the x divided by three to the X minus one. And we see that when we plug in zero we get 0/0. So because this is an interment form um we're not going to be allowed to do this directly. We're going to have to use low towels will instead. So what that's gonna look like is um this could be rewritten as X over one minus one third ax power. Um And then with this in mind we now can take the limit as X goes to zero. So this is going to end up giving us as a result one as the numerator, this will give us one and then as the denominator this would go away and we'd be left with um one over negative Natural Log of 1/3. Yeah so with that we see that that's what happens when we multiply. Or when we plug in the zero value negative natural log of 1/3 is just the natural log of three. So this will be our final answer.