જો f^{prime prime}(0)=k, k neq 0, હોય, તો lim | vedclass

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(C) આપેલ લક્ષ $L = \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$.લક્ષ $\frac{0}{0}$ સ્વરૂપમાં હોવાથી, આપણે $L$'$H$ôpital ના નિયમનો ઉપય

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(C) આપેલ લક્ષ $L = \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$.લક્ષ $\frac{0}{0}$ સ્વરૂપમાં હોવાથી, આપણે $L$'$H$ôpital ના નિયમનો ઉપયોગ કરીએ:$L = \lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-6 f^{\prime}(2 x)+4 f^{\prime}(4 x)}{2 x}$$L = \lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$ફરીથી $L$'$H$ôpital નો નિયમ વાપરતા:$L = \lim _{x}$ ${\rightarrow 0} \frac{f^{\prime \prime}(x)-3 f^{\prime \prime}(2 x) \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$$L = f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f^{\prime \prime}(0)$$L = k-6 k+8 k = 3 k$.

જો $f^{\prime \prime}(0)=k, k \neq 0,$ હોય, તો $\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$ ની કિંમત શું થાય?

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