A) 0
B) \[-1/6\]
C) \[1/6\]
D) Does not exist
Correct Answer: C
Solution :
| [c] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{4}}-{{x}^{4}}\cos {{x}^{4}}+{{x}^{20}}}{{{x}^{4}}({{e}^{2{{x}^{4}}}}1-2{{x}^{4}})}\] |
| \[=\underset{t\to 0}{\mathop{\lim }}\,\frac{\sin t-t\cos t+{{t}^{5}}}{t({{e}^{2t}}-1-2t)}\] |
| \[=\underset{t\to 0}{\mathop{\lim }}\,\frac{t-\frac{{{t}^{3}}}{3!}+\frac{{{t}^{5}}}{5!}...-t\left( 1-\frac{{{t}^{2}}}{2!}+\frac{{{t}^{4}}}{4!}... \right)+{{t}^{5}}}{t\left( 1+2t+\frac{4{{t}^{2}}}{2!}+\frac{8{{t}^{3}}}{3!}+\frac{16{{t}^{4}}}{4!}+...-1-2t \right)}\] |
| \[=\underset{t\to 0}{\mathop{\lim }}\,\frac{-\frac{{{t}^{3}}}{6}+\frac{{{t}^{3}}}{2}+\frac{{{t}^{5}}}{5!}-\frac{{{t}^{5}}}{4!}+...+{{t}^{5}}}{2{{t}^{3}}+\frac{8{{t}^{4}}}{3!}+....}\] |
| \[=\frac{-\frac{1}{6}+\frac{1}{2}}{2}=-\frac{-1+3}{12}=\frac{1}{6}\] |
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