f(x) is differentiable on mathbb{R} and f^{pr | vedclass

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(C) Given the limit expression: $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$.We can rewrite the numerator by adding and subt

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$1$

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(C) Given the limit expression: $\lim _{x ightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$.We can rewrite the numerator by adding and subtracting $m f(m)$:$\lim _{x ightarrow m} \frac{x f(m)-m f(m)+m f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$$\lim _{x ightarrow m} \left[ f(m) \frac{x-m}{x-m} - m \frac{f(x)-f(m)}{x-m} ight] + f^{\prime}(m) = f(m)$$f(m) - m f^{\prime}(m) + f^{\prime}(m) = f(m)$$-m f^{\prime}(m) + f^{\prime}(m) = 0$$f^{\prime}(m)(1-m) = 0$Since it is given that $f^{\prime}(m) 
eq 0$,we must have $1-m = 0$,which implies $m = 1$.

$f(x)$ is differentiable on $\mathbb{R}$ and $f^{\prime}(m) \neq 0, \,m \in \mathbb{R}$. If $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$,then $m=$

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