Question

\(\begin{array}{l}\displaystyle \lim_{ x\to 0}\left(\frac{(x+2\cos x)^3 + 2(x+2\cos x)^2 + 3 \sin(x+2\cos x)}{(x+2)^3+2(x+2)^2+3\sin(x+2)}\right)^{\frac{100}{x}} \end{array}\)

is equal to _________.

Answer 1

The correct answer is 1
Let,

\(x + 2cosx = a\)

\(x + 2 = b\)

\(as \space x → 0, a → 2 \space and \space b → 2\)

\(\begin{array}{l}\displaystyle \lim_{ x\to 0}\left(\frac{a^3+2a^2+3\sin a}{b^3 + 2b^2 + 3\sin b}\right)^{\frac{100}{x}} \end{array}\)

\(\begin{array}{l}=e^{\displaystyle \lim_{x \to 0}\frac{100}{x}.\frac{(a^3-b^3)+2(a^2-b^2)+3(\sin a – \sin b)}{b^3+2b^2 + 3\sin b}}\end{array}\)

\(\begin{array}{l}\because \displaystyle \lim_{x\to 0}\frac{a-b}{x}=\displaystyle \lim_{x\to 0}\frac{2(\cos x -1)}{x}=0\end{array}\)

\(= e^0\)

\(= 1\)