Let f: R -> (0, infinity) be a twice differentiable function such that f(3)=18, f^prime(3)=0 and f^prime prime(3)=4.\ Then lim _x -> 1(log _e((f(2+x) / f(3)))^frac18(x-1)^2) is equal to :

Let T=lim _x -> 1((f(x+2) / f(3)))^frac18(x-1)^2 ; 1^infinity form \ => T= e^lim _ x -> 1 frac18( x-1)^2(frac f( x+2)- f(3) f(3)) => T= e^lim _ x -> 1 frac18( x-1)^2(frac f( x+2)- f(3)18) => T= e^lim _ x -> 1(frac f( x+2)- f(3)( x-1)^2) (0 / 0) form apply L' pital \ => T= e^lim _ x -> 1 frac f^prime( x+2)2( x-1) ; (0 / 0) form apply L' pital \ => T= e^lim _ x -> 1 frac f^prime prime( x+2)2 ;= e^(4 / 2)= e^2 => log _ e( T)=2

Let f R arrow 0 infty be a twice differentiable functio... | null (Mathematics)

Let

T=\lim _{x \rightarrow 1}\left(\frac{f(x+2)}{f(3)}\right)^{\frac{18}{(x-1)^{2}}} ; 1^{\infty}

form

\\ \Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{18}{(\mathrm{x}-1)^{2}}\left(\frac{\mathrm{f}(\mathrm{x}+2)-\mathrm{f}(3)}{\mathrm{f}(3)}\right)}\\

\Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{18}{(\mathrm{x}-1)^{2}}\left(\frac{\mathrm{f}(\mathrm{x}+2)-\mathrm{f}(3)}{18}\right)}\\

\Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1}\left(\frac{\mathrm{f}(\mathrm{x}+2)-\mathrm{f}(3)}{(\mathrm{x}-1)^{2}}\right) \frac{0}{0} \text { form }}

apply L' pital

\\ \Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}^{\prime}(\mathrm{x}+2)}{2(\mathrm{x}-1)} ; \frac{0}{0} \text { form }} \quad

apply L' pital

\\ \Rightarrow \mathrm{T}=\mathrm{e}^{\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}^{\prime \prime}(\mathrm{x}+2)}{2}} ;=\mathrm{e}^{\frac{4}{2}}=\mathrm{e}^{2}\\

\Rightarrow \log _{\mathrm{e}}(\mathrm{T})=2

Explanation

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