A) (a) 0
B) (b) 1
C) (c) \[\frac{1}{6}\]
D) (d) \[\frac{-1}{6}\]
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}+{{e}^{-x}}+2.cosx-4}{{{x}^{4}}}\] Which is \[\frac{0}{0}\] form \[=\underset{x\to 0}{\mathop{lim}}\,\frac{{{e}^{x}}-{{e}^{-x}}-2.\sin x}{4{{x}^{3}}}\] [By L Hospital rules] \[=\underset{x\to 0}{\mathop{lim}}\,{{e}^{x}}\]which is\[\frac{0}{0}\]form. \[=\underset{x\to 0}{\mathop{lim}}\,\frac{{{e}^{x}}+{{e}^{-x}}-2.\cos x}{12{{x}^{2}}}\][By L. Hospital rule] Again which is \[\frac{0}{0}\]form \[=\underset{x\to 0}{\mathop{lim}}\,\frac{{{e}^{x}}-{{e}^{-x}}-2.\sin x}{24x}=\frac{1+1+2}{24}=\frac{1}{6}\] [which is\[\frac{0}{0}\]form] \[=\underset{x\to 0}{\mathop{lim}}\,\frac{{{e}^{x}}+{{e}^{-x}}+2.\cos x}{24}=\frac{1+1+2}{24}=\frac{1}{6}\] Hence, option [c] is correct.
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