If lim _{x rightarrow 0} frac{e^x-a-log (1+x) | vedclass

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(D) Given $\lim _{x \rightarrow 0} \frac{e^x-a-\log (1+x)}{\sin x}=0$. For the limit to exist and be finite,the numerator must approach $0$ as $x \rig

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(D) Given $\lim _{x \rightarrow 0} \frac{e^x-a-\log (1+x)}{\sin x}=0$. For the limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$ because the denominator $\sin x \rightarrow 0$. Substituting $x=0$ in the numerator: $e^0 - a - \log(1+0) = 0$. $1 - a - 0 = 0$. Therefore,$a = 1$. Checking with $a=1$: $\lim _{x \rightarrow 0} \frac{e^x-1-\log (1+x)}{\sin x}$. Using $L$'Hopital's rule: $\lim _{x \rightarrow 0} \frac{e^x - \frac{1}{1+x}}{\cos x} = \frac{1-1}{1} = 0$. Thus,the condition holds for $a=1$.

If $\lim _{x \rightarrow 0} \frac{e^x-a-\log (1+x)}{\sin x}=0$,then $a=$

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