If f(x)=3 x^{15}-5 x^{10}+7 x^5+50 cos (x-1), | vedclass

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(C) Given $f(x)=3 x^{15}-5 x^{10}+7 x^5+50 \cos (x-1)$.We need to evaluate $L = \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^3+3 h}$.Rewrite the limit

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-$10$

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(C) Given $f(x)=3 x^{15}-5 x^{10}+7 x^5+50 \cos (x-1)$.We need to evaluate $L = \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^3+3 h}$.Rewrite the limit as $L = \lim _{h \rightarrow 0} \left( \frac{f(1-h)-f(1)}{-h} \cdot \frac{-h}{h(h^2+3)} \right)$.Since $\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} = f'(1)$,we have $L = f'(1) \cdot \lim _{h \rightarrow 0} \left( \frac{-1}{h^2+3} \right)$.Now,$f'(x) = 45 x^{14} - 50 x^9 + 35 x^4 - 50 \sin (x-1)$.Evaluating at $x=1$,$f'(1) = 45(1)^{14} - 50(1)^9 + 35(1)^4 - 50 \sin(0) = 45 - 50 + 35 = 30$.Thus,$L = 30 \cdot \left( \frac{-1}{0^2+3} \right) = 30 \cdot \left( -\frac{1}{3} \right) = -10$.

If $f(x)=3 x^{15}-5 x^{10}+7 x^5+50 \cos (x-1)$,then $\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^3+3 h}=$

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