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# Illustrate l'Hospital's Rule by graphing both $f(x)/g(x)$ and $f'(x)/g'(x)$ near $x = 0$ to see that these ratios have the same limit as $x \to 0$. Also, calculate the exact value of the limit.$f(x) = e^x - 1$, $g(x) = x^3 + 4x$

## From the graph, it appears that $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}=0.25$$\text { We calculate } \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{e^{x}-1}{x^{3}+4 x} \stackrel{\text { ? }}{=} \lim _{x \rightarrow 0} \frac{e^{x}}{3 x^{2}+4}=\frac{1}{4}$

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### Video Transcript

mhm. This problem we're going to be looking at the graph as edX -1. Okay. Divided by X cubed. That's four X. So this is the graph we get um and then that's fx and gx but then F prime of X is going to be needy. X and G prime of X is going to be three X squared plus four. Um So going back, if we autographed separately, we have each of the acts divided by three X squared response for what we see. Is that the green curve? Um If we zoom in both curves approached the same number at X equal to zero And that value appears to be .25. Um And if we calculate the limit we see that that in fact is the case Because when we plug in zero here we get one over four Us, that would be .25.

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