If lim _{x rightarrow 0} frac{a x e^{x}-b log | vedclass

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(A) Given $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$. Since the limit exists and the denominator approaches $0$,the numerator mu

Vedclass · Accepted Answer

$2, 2$

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(A) Given $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$. Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ as $x \rightarrow 0$.Applying $L$'Hospital's rule once: $\lim _{x \rightarrow 0} \frac{a e^{x} + a x e^{x} - \frac{b}{1+x}}{2x} = 3$.For the limit to exist,the numerator must be $0$ at $x=0$: $a(1) + a(0) - b = 0$ $\Rightarrow a - b = 0$ $\Rightarrow a = b$.Applying $L$'Hospital's rule again: $\lim _{x \rightarrow 0} \frac{a e^{x} + a e^{x} + a x e^{x} + \frac{b}{(1+x)^{2}}}{2} = 3$.Substituting $x=0$: $\frac{a + a + 0 + b}{2} = 3 \Rightarrow 2a + b = 6$.Since $a = b$,we have $2a + a = 6$ $\Rightarrow 3a = 6$ $\Rightarrow a = 2$.Thus,$b = 2$.

If $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$,then the values of $a$ and $b$ are,respectively:

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