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If $f(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7$, then $\lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}$ is
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Verified Answer
The correct answer is:
$\frac{53}{3}$
$\frac{53}{3}$
$$
\text { } \begin{aligned}
\text { Let } f^{\prime}(x) & =3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7 \\
f^{\prime}(x) & =30 x^9-56 x^7+30 x^5-63 x^2+6 x \\
f^{\prime}(1) & =30-56+30-63+6 \\
& =66-63-56=-53
\end{aligned}
$$
$$
\begin{aligned}
& \text { Consider } \lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha} \\
& =\lim _{\alpha \rightarrow 0} \frac{f^{\prime}(1-\alpha)(-1)-0}{3 \alpha^2+3} \quad \text { (By using }
\end{aligned}
$$
L'hospital rule)
$$
=\frac{f^{\prime}(1-0)(-1)}{3(0)^2+3}=\frac{-f^{\prime}(1)}{3}=\frac{53}{3}
$$
\text { } \begin{aligned}
\text { Let } f^{\prime}(x) & =3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7 \\
f^{\prime}(x) & =30 x^9-56 x^7+30 x^5-63 x^2+6 x \\
f^{\prime}(1) & =30-56+30-63+6 \\
& =66-63-56=-53
\end{aligned}
$$
$$
\begin{aligned}
& \text { Consider } \lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha} \\
& =\lim _{\alpha \rightarrow 0} \frac{f^{\prime}(1-\alpha)(-1)-0}{3 \alpha^2+3} \quad \text { (By using }
\end{aligned}
$$
L'hospital rule)
$$
=\frac{f^{\prime}(1-0)(-1)}{3(0)^2+3}=\frac{-f^{\prime}(1)}{3}=\frac{53}{3}
$$
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