Let l = mathop {Lim}limits_{x to infty } intl | vedclass

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(A) First,evaluate $l$: $l = \mathop {Lim}\limits_{x 	o \infty } [\ln t]_x^{2x} = \mathop {Lim}\limits_{x 	o \infty } (\ln(2x) - \ln x) = \mathop {L

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$l \cdot m = l$

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(A) First,evaluate $l$: $l = \mathop {Lim}\limits_{x 	o \infty } [\ln t]_x^{2x} = \mathop {Lim}\limits_{x 	o \infty } (\ln(2x) - \ln x) = \mathop {Lim}\limits_{x 	o \infty } \ln(\frac{2x}{x}) = \ln 2$. Next,evaluate $m$: $m = \mathop {Lim}\limits_{x 	o \infty } \frac{\int_1^x \ln t \, dt}{x \ln x}$. Using $L$'Hopital's Rule,differentiate the numerator and denominator with respect to $x$: Numerator: $\frac{d}{dx} \int_1^x \ln t \, dt = \ln x$. Denominator: $\frac{d}{dx} (x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$. Thus,$m = \mathop {Lim}\limits_{x 	o \infty } \frac{\ln x}{\ln x + 1} = \mathop {Lim}\limits_{x 	o \infty } \frac{1}{1 + \frac{1}{\ln x}} = \frac{1}{1 + 0} = 1$. Therefore,$l \cdot m = \ln 2 \cdot 1 = \ln 2 = l$.

Let $l = \mathop {Lim}\limits_{x \to \infty } \int\limits_x^{2x} \frac{dt}{t}$ and $m = \mathop {Lim}\limits_{x \to \infty } \frac{1}{x \ln x} \int\limits_1^x \ln t \, dt$. Then the correct statement is:

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