If mathop {lim }limits_{x to 0} frac{{ln left | vedclass

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(A) We use the Taylor series expansion for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ $\mathop {\lim }\limits_{x 	o 0} \frac{{\left( {x -

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$\frac{1}{2}$

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(A) We use the Taylor series expansion for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ $\mathop {\lim }\limits_{x 	o 0} \frac{{\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \dots} ight) - ax}}{{{x^2}}} = l$ $= \mathop {\lim }\limits_{x 	o 0} \frac{{\left( {1 - a} ight)x - \frac{1}{2}{x^2} + \frac{{{x^3}}}{3} - \dots}}{{{x^2}}} = l$ For the limit to be a finite number $l$,the coefficient of $x$ must be zero. Therefore,$1 - a = 0 \implies a = 1$. Substituting $a = 1$,we get $l = \mathop {\lim }\limits_{x 	o 0} \frac{{ - \frac{1}{2}{x^2} + \frac{{{x^3}}}{3}}}{{{x^2}}} = -\frac{1}{2}$. Thus,$a + l = 1 + (-\frac{1}{2}) = \frac{1}{2}$.

If $\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right) - ax}}{{{x^2}}} = l$,then the value of $(a + l)$ is equal to (where $l$ is a finite number).

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