Question

The limit $ \displaystyle \lim_{x \to {{0}^{+}}}\left( {{\left( x\cos x \right)}^{x}}+{{\left( cosecx \right)}^{\dfrac{1}{\ln x}}}+{{\left( x\sin x \right)}^{x}} \right) $ is equal to \[\]
A.2\[\]
B. $ 2+e $ \[\]
C. $ 2+\dfrac{1}{e} $ \[\]
D.3\[\]

Answer 1

Hint: We recall use sum rule of limits to separately evaluate $ \displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\cos x \right)}^{x}},\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\sin x \right)}^{x}} $ and $ \displaystyle \lim_{x \to {{0}^{+}}}{{\left( \cos x \right)}^{\dfrac{1}{\ln x}}} $ . We evaluate the $ \displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\cos x \right)}^{x}},\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\sin x \right)}^{x}} $ by using definition of logarithm $ {{e}^{\ln a}}=a $ . We evaluate $ \displaystyle \lim_{x \to {{0}^{+}}}{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{\ln x}}} $ by taking $ \operatorname{cosec}x=\dfrac{1}{\sin x} $ and then taking logarithm both side of $ y=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{\ln x}}} $ and then using L’hospital rule for $ \dfrac{\infty }{\infty } $ form. \[\]

Complete step by step answer:
We know that we can use L’hospital rule to evaluate limits if the function is in the indeterminate forms like $ \dfrac{0}{0},\dfrac{\infty }{\infty },0\times \infty ,\infty -\infty $ etc. The L’hospital rule is stated for $ \dfrac{\infty }{\infty } $ is stated as follows: if $ \displaystyle \lim_{x \to c}f\left( x \right)=\infty ,\displaystyle \lim_{x \to c}g\left( x \right)=\infty $ , $ {{g}^{'}}\left( x \right)\ne 0 $ for all $ x\ne c $ in the domain and $ \dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)} $ exist then
\[\displaystyle \lim_{x \to c}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to c}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}\]
Let us evaluate the limit $ \displaystyle \lim_{x \to {{0}^{+}}}x\ln x $ using l’hospital rule for indeterminate form $ \dfrac{\infty }{\infty } $ which we are going to use later. We have;
\[\displaystyle \lim_{x \to {{0}^{+}}}x\ln x=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\ln x}{\dfrac{1}{x}}=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\dfrac{1}{x}}{-\dfrac{1}{{{x}^{2}}}}=\displaystyle \lim_{x \to {{0}^{+}}}\left( -x \right)=0\]
We are given the following right hand limit to evaluate at $ x=0 $ ,
\[\displaystyle \lim_{x \to {{0}^{+}}}\left( {{\left( x\cos x \right)}^{x}}+{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{\ln x}}}+{{\left( x\sin x \right)}^{x}} \right)\]
We use law of addition of limits and have;
\[\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\cos x \right)}^{x}}+\displaystyle \lim_{x \to {{0}^{+}}}{{\left( cosecx \right)}^{\dfrac{1}{\ln x}}}+\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\sin x \right)}^{x}}={{L}_{1}}+{{L}_{2}}+{{L}_{3}}\left( \text{say} \right)\]
Let us evaluate the first limit $ {{L}_{1}}=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\cos x \right)}^{x}} $ by using exponential identity $ {{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}} $ for $ a=x,b=\cos x,m=x $ to have;
\[\begin{align}
  & {{L}_{1}}=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\cos x \right)}^{x}} \\
 & \Rightarrow {{L}_{1}}=\displaystyle \lim_{x \to {{0}^{+}}}{{x}^{x}}{{\left( \cos x \right)}^{x}} \\
\end{align}\]
We use the definition of logarithm $ {{e}^{\ln a}}=a $ for $ a={{x}^{x}}\Rightarrow \ln a=x\ln x $ in the above step to have;
\[\Rightarrow {{L}_{1}}=\displaystyle \lim_{x \to {{0}^{+}}}{{e}^{x\ln x}}{{\left( \cos x \right)}^{x}}\]
We use law of multiplication of limits and put previously obtained $ \displaystyle \lim_{x \to {{0}^{+}}}x\ln x=0 $ to have;
\[\begin{align}
  & \Rightarrow {{L}_{1}}={{e}^{\displaystyle \lim_{x \to {{0}^{+}}}x\ln x}}\cdot \displaystyle \lim_{x \to {{0}^{+}}}{{\left( \cos x \right)}^{x}} \\
 & \Rightarrow {{L}_{1}}={{e}^{0}}\cdot {{\left( \cos 0 \right)}^{0}} \\
 & \Rightarrow {{L}_{1}}==1\cdot {{1}^{0}}=1\cdot 1=1 \\
\end{align}\]
Similarly we evaluate $ {{L}_{3}}=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\sin x \right)}^{x}} $ by using exponential identity $ {{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}} $ for $ a=x,b=\sin x,m=x $ to have;
 \[\begin{align}
  & {{L}_{3}}=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\sin x \right)}^{x}} \\
 & \Rightarrow {{L}_{3}}=\displaystyle \lim_{x \to {{0}^{+}}}{{x}^{x}}{{\left( \sin x \right)}^{x}} \\
\end{align}\]
We use law of multiplication of limits and have;
 $ \Rightarrow {{L}_{3}}=\displaystyle \lim_{x \to {{0}^{+}}}{{x}^{x}}\cdot \displaystyle \lim_{x \to {{0}^{+}}}{{\left( \sin x \right)}^{x}} $
We use the definition of logarithm $ {{e}^{\ln a}}=a $ for $ a={{x}^{x}}\Rightarrow \ln a=x\ln x $ for the first limit and then $ a={{\left( \sin x \right)}^{x}}\Rightarrow \ln a=x\ln \sin x $ for in the above step to have;
 \[\begin{align}
  & \Rightarrow {{L}_{3}}=\displaystyle \lim_{x \to {{0}^{+}}}{{e}^{x\ln x}}\cdot \displaystyle \lim_{x \to {{0}^{+}}}{{e}^{x\ln \sin x}} \\
 & \Rightarrow {{L}_{3}}={{e}^{\displaystyle \lim_{x \to {{0}^{+}}}x\ln x}}\cdot {{e}^{\displaystyle \lim_{x \to {{0}^{+}}}x\ln \sin x}} \\
\end{align}\]
We use put previously obtained $ \displaystyle \lim_{x \to {{0}^{+}}}x\ln x=0 $
\[\begin{align}
  & \Rightarrow {{L}_{3}}=1\cdot {{e}^{\displaystyle \lim_{x \to {{0}^{+}}}x\ln \sin x}} \\
 & \Rightarrow {{L}_{3}}={{e}^{\displaystyle \lim_{x \to {{0}^{+}}}x\ln \sin x}} \\
 & \Rightarrow {{L}_{3}}={{e}^{\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\ln \sin x}{\dfrac{1}{x}}}} \\
\end{align}\]
We use l’hospital rule for indeterminate form $ \dfrac{\infty }{\infty } $ and differentiate the numerator and denominator to have;
\[\begin{align}
  & \Rightarrow {{L}_{3}}={{e}^{\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\cos x\cdot \dfrac{1}{\sin x}}{-\dfrac{1}{{{x}^{2}}}}}} \\
 & \Rightarrow {{L}_{3}}={{e}^{\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{-{{x}^{2}}\cos x}{\sin x}}} \\
\end{align}\]
We again use l’hospital rule for indeterminate form $ \dfrac{0}{0} $ and differentiate the numerator and denominator to have;
 \[\begin{align}
  & \Rightarrow {{L}_{3}}={{e}^{\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{{{x}^{2}}\sin x-2x\cos x}{\cos x}}} \\
 & \Rightarrow {{L}_{3}}={{e}^{\dfrac{0}{1}}}={{e}^{0}}=1 \\
\end{align}\]
Now let us evaluate $ {{L}_{2}}=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( cosecx \right)}^{\dfrac{1}{\ln x}}} $ . Let us have $ y=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{\ln x}}} $ and then use reciprocal relation between sine and cosecant that is $ \operatorname{cosec}\theta =\dfrac{1}{\sin \theta } $ to have;
\[y=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( \dfrac{1}{\sin x} \right)}^{\dfrac{1}{\ln x}}}\]
We take logarithm both sides to have;
\[\begin{align}
  & \ln y=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{\ln x}\ln \left( \dfrac{1}{\sin x} \right) \\
 & \Rightarrow \ln y=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{\ln x}\left( \ln 1-\ln \sin x \right) \\
 & \Rightarrow \ln y=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{1}{\ln x}\left( -\ln \sin x \right)\left( \because \ln 1=0 \right) \\
 & \Rightarrow \ln y=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{-\ln \sin x}{\ln x} \\
\end{align}\]
We use l’hospital rule for indeterminate form $ \dfrac{\infty }{\infty } $ and differentiate the numerator and denominator to have;

\[\begin{align}
  & \Rightarrow \ln y=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\cos x\cdot \dfrac{-1}{\sin x}}{\dfrac{1}{x}} \\
 & \Rightarrow \ln y=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{-x\cos x}{\sin x} \\
\end{align}\]
We again use l’hospital rule for indeterminate form $ \dfrac{0}{0} $ and differentiate the numerator and denominator to have;

\[\begin{align}
  & \Rightarrow \ln y=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{x\sin x-\cos x}{\cos x} \\
 & \Rightarrow \ln y=\dfrac{0\cdot 0-1}{1} \\
 & \Rightarrow \ln y=-1 \\
\end{align}\]
We use definition of natural logarithm to have;
\[\begin{align}
  & \Rightarrow y={{e}^{-1}} \\
 & \Rightarrow \displaystyle \lim_{x \to {{0}^{+}}}{{\left( cosecx \right)}^{\dfrac{1}{\ln x}}}={{e}^{-1}} \\
 & \Rightarrow {{L}_{2}}=\dfrac{1}{e} \\
\end{align}\]
So the evaluated limit is
\[\begin{align}
  & \displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\cos x \right)}^{x}}+\displaystyle \lim_{x \to {{0}^{+}}}{{\left( cosecx \right)}^{\dfrac{1}{\ln x}}}+\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\sin x \right)}^{x}}={{L}_{1}}+{{L}_{2}}+{{L}_{3}} \\
 & \Rightarrow \displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\cos x \right)}^{x}}+\displaystyle \lim_{x \to {{0}^{+}}}{{\left( cosecx \right)}^{\dfrac{1}{\ln x}}}+\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x\sin x \right)}^{x}}=1+\dfrac{1}{e}+1=2+\dfrac{1}{e} \\
\end{align}\]
So the correct choice is C.\[\]


Note:
 We note that since we are asked to evaluate the right hand limit that is $ x>0 $ , the function $ \ln x $ is well defined and $ \displaystyle \lim_{x \to {{0}^{+}}}\sin x=\sin x,\displaystyle \lim_{x \to {{0}^{+}}}\cos x=\cos x, $ We know from the laws of limits for the limit for two real valued functions $ f\left( x \right) $ and $ g\left( x \right) $ at point $ x=a $ for both functions then by law of addition in limits we have $ \displaystyle \lim_{x \to a}f\left( x \right)+\displaystyle \lim_{x \to a}g\left( x \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right)+g\left( x \right) \right) $ and by law of multiplication of limits $ \displaystyle \lim_{x \to a}f\left( x \right)\cdot \displaystyle \lim_{x \to a}g\left( x \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right)\cdot g\left( x \right) \right) $ . We should remember the logarithmic identities of quotient $ \log \left( \dfrac{a}{b} \right)=\log a-\log b $ .