If limx→−1 sin⁡x3+bx2+cx+d(2+x−1)loga⁡(x+2)2 exists and Is equal to I, then b + d + I is equal to

Solution:

It is given that

$\begin{array}{l} lim_{x \to - 1} \frac{\sin (x^{3} + b x^{2} + c x + d)}{(\sqrt{2 + x} - 1) {\{\log (x + 2)\}}^{2}} = l \\ \Rightarrow lim_{x \to - 1} \frac{\sin (x^{3} + b x^{2} + c x + d)}{(x + 1)^{3} {\{\log_{e} \frac{(1 + (x + 1))}{(x + 1)}\}}^{2} \times (\sqrt{2 + x} + 1)} = l \\ \Rightarrow lim_{x \to - 1} \frac{\sin (x^{3} + b x^{2} + c x + d)}{(x^{3} + b x^{2} + c x + d)} \times \frac{1}{{\{\frac{\log_{e} (1 + (x + 1))}{x + 1}\}}^{2}} \times \frac{x^{3} + b x^{2} + c x + d}{(x + 1)^{3}} \times (\sqrt{2 + x} + 1) = 1 \end{array}$

$\begin{array}{l} \Rightarrow lim_{x \to - 1} 1 \times \frac{1}{(1)^{2}} \times \frac{x^{3} + b x^{2} + c x + d}{(x + 1)^{3}} \times 2 = l \\ \Rightarrow lim_{x \to - 1} \frac{x^{3} + b x^{2} + c x + d}{(x + 1)^{3}} = \frac{1}{2} \\ \Rightarrow b = 3, c = 3, d = 1 and l = 2 \\ ∴ b + d + l = 3 + 1 + 2 = 6 \end{array}$

If $lim_{x \to - 1} \frac{\sin (x^{3} + b x^{2} + c x + d)}{(\sqrt{2 + x} - 1) {\{\log_{a} (x + 2)\}}^{2}}$ exists and Is equal to I, then $b + d + I$ is equal to

Solution:

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