If lim _{x rightarrow 0} frac{e^{2 x}-1}{5 x} | vedclass

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(C) First,evaluate $l$: $\lim _{x ightarrow 0} \frac{e^{2 x}-1}{5 x} = \frac{1}{5} \lim _{x ightarrow 0} \frac{e^{2 x}-1}{x} = \frac{1}{5} 	imes

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$x^3-5 x^2+8 x-4=0$

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(C) First,evaluate $l$: $\lim _{x ightarrow 0} \frac{e^{2 x}-1}{5 x} = \frac{1}{5} \lim _{x ightarrow 0} \frac{e^{2 x}-1}{x} = \frac{1}{5} 	imes 2 = \frac{2}{5}$. So,$l = \frac{2}{5}$. Then,$5l = 5 	imes \frac{2}{5} = 2$. Next,evaluate $m$: $\lim _{x ightarrow 1} \frac{2 \log x}{x-1}$. Let $x = 1+h$,as $x ightarrow 1, h ightarrow 0$. $\lim _{h ightarrow 0} \frac{2 \log(1+h)}{h} = 2(1) = 2$. So,$m = 2$. The roots of the cubic equation are $2, 2$,and $1$. The equation is $(x-2)(x-2)(x-1) = 0$. $(x^2-4x+4)(x-1) = 0$. $x^3 - x^2 - 4x^2 + 4x + 4x - 4 = 0$. $x^3 - 5x^2 + 8x - 4 = 0$.

If $\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{5 x}=l$ and $\lim _{x \rightarrow 1} \frac{2}{x-1} \log x=m$,then the cubic equation whose roots are $5l, m$,and $1$ is:

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