A) 2
B) 2 + e
C) \[2+\frac{1}{e}\]
D) 3
Correct Answer: C
Solution :
\[l=\underset{x\to {{o}^{+}}}{\mathop{\lim }}\,\left( {{(x\cos x)}^{2}}+{{(\cos ecx)}^{\frac{1}{\ln x}}}+{{(x\sin x)}^{x}} \right)\] \[l=\underbrace{\underset{x\to 0}{\mathop{\text{L}im}}\,{{(x\cos x)}^{x}}}_{{{l}_{1}}}+l=\underbrace{\underset{x\to {{0}^{+}}}{\mathop{\text{L}im}}\,{{(x\cos ec\,x)}^{\frac{1}{x}}}}_{{{l}_{2}}}+\underbrace{\underset{x\to {{0}^{+}}}{\mathop{\text{L}im}}\,{{(x\sin x)}^{x}}}_{{{l}_{3}}}\]\[{{l}_{1}}={{(x\cos x)}^{x}}\] \[\ln {{l}_{1}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (x\cos x)}{\left( \frac{1}{x} \right)}=\frac{-x\sin x+\cos x}{x\cos x\left( \frac{-1}{{{x}^{2}}} \right)}\] \[=0\Rightarrow {{l}_{1}}=1\] \[|{{|}^{ly}}{{l}_{2}}=\frac{1}{e}\]and\[{{l}_{3}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
