If \[f\left( x \right)=\left\{ \begin{align}
& \dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left[ x+1 \right]}}-9}{{{3}^{x}}-27}\text{ ; }x<3 \\
& \lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}\text{ ; }x>3 \\
\end{align} \right.\] is continuous at \[x=3\], then the value of \[8100\lambda \] must be?
Answer
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Hint: In order to find the solution of the given question, that is to determine the value of \[8100\lambda \] if \[f\left( x \right)=\left\{ \begin{align}
& \dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left[ x+1 \right]}}-9}{{{3}^{x}}-27}\text{ ; }x<3 \\
& \lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}\text{ ; }x>3 \\
\end{align} \right.\] is continuous at \[x=3\]. Apply the one of the results of continuous function that is “if a function \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\] “to find the value of \[\lambda \].
Complete step by step answer:
According to the question, given function in question is as follows:
\[f\left( x \right)=\left\{ \begin{align}
& \dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left[ x+1 \right]}}-9}{{{3}^{x}}-27}\text{ ; }x<3 \\
& \lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}\text{ ; }x>3 \\
\end{align} \right.\]
We know that this function is continuous at \[x=3\]. Therefore, according to the definition of continuous function that is if a function \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\] .
This implies that as per the given in the question, \[f\left( x \right)\] is continuous at \[x=3\] then \[\displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)\]
Now, let’s first find the limit of the left-hand side of the function.
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( {{e}^{\left( \left( x+3 \right)\ln 27 \right)}} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
Apply the property of logarithm that is \[\left( a \right)\ln \left( b \right)=\ln {{\left( b \right)}^{a}}\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( {{e}^{\left( \ln {{27}^{\left( x+3 \right)}} \right)}} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
We know that \[{{e}^{\ln a}}=a\], therefore we can rewrite above expression as follows:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( {{27}^{\left( x+3 \right)}} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( 27 \right)}^{\dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
Now take the term \[9\] common from the numerator of the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{9.\left( {{27}^{\dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3}}}-1 \right)}{27\left( \dfrac{{{3}^{x}}}{27}-1 \right)}\]
Now take the term \[27\] common from the denominator of the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{9.\left( {{27}^{\dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3}}}-1 \right)}{27\left( {{3}^{x-3}}-1 \right)}\]
After this multiply and divide term \[\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( {{27}^{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}}-1 \right)}{3\left( {{3}^{x-3}}-1 \right)}\times \dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}\]
Apply the result that \[\displaystyle \lim_{x \to b}\dfrac{{{a}^{x}}-1}{x}=1\], therefore we can say that \[\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( {{27}^{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}}-1 \right)}{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}=1\], hence, we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{3\left( {{3}^{x-3}}-1 \right)}\]
After this multiply and divide term \[\left( x-3 \right)\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{3\left( {{3}^{\left( x-3 \right)}}-1 \right)\times \dfrac{\left( x-3 \right)}{\left( x-3 \right)}}\]
Apply the result that \[\displaystyle \lim_{x \to b}\dfrac{{{a}^{x}}-1}{x}=1\], therefore we can say that \[\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( {{3}^{\left( x-3 \right)}}-1 \right)}{\left( x-3 \right)}=1\], hence, we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{3\left( x-3 \right)}\]
Now, take the LCM of the terms in the numerator, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{\left( x+3 \right)\left( x+1 \right)-18}{27} \right)}{3\left( x-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( x+3 \right)\left( x+1 \right)-18}{3\times 27\times \left( x-3 \right)}\]
After simplifying this further we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{x}^{2}}+x+3x+3-18}{3\times 27\times \left( x-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{x}^{2}}+4x-15}{81\left( x-3 \right)}\]
Now, apply the left-hand side limit that is putting \[x=3-h\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{{{\left( 3-h \right)}^{2}}+4\left( 3-h \right)-15}{81\left( 3-h-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{{{\left( 3-h \right)}^{2}}+4\left( 3-h \right)-15}{81\left( -h \right)}\]
\[\Rightarrow \dfrac{{{\left( 3-0 \right)}^{2}}+4\left( 3-0 \right)-15}{81\left( -0 \right)}\]
\[\Rightarrow -\infty \]
Therefore, left-hand side limit of the function is equal to \[-\infty \]which means that\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}=-\infty \]
Now find the right-hand side limit of the function.
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}\]
Apply the identity that is \[1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}\]in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{2{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\left( x-3 \right)\tan \left( x-3 \right)}\]
Now multiply and divide \[\dfrac{\left( x-3 \right)}{4}\]in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{\dfrac{\left( x-3 \right)}{4}}{\dfrac{\left( x-3 \right)}{4}}\dfrac{2{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\left( x-3 \right)\tan \left( x-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{x \to +3}\lambda .\dfrac{\left( x-3 \right)}{4}.\dfrac{2{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\dfrac{{{\left( x-3 \right)}^{2}}}{4}\tan \left( x-3 \right)}\]
Apply the result that \[\displaystyle \lim_{x \to b}\dfrac{{{\sin }^{2}}x}{{{x}^{2}}}=1\], therefore we can say that \[\displaystyle \lim_{x \to {{3}^{+}}}\dfrac{{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\dfrac{{{\left( x-3 \right)}^{2}}}{4}}=1\], hence, we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\dfrac{\left( x-3 \right)}{4}.\dfrac{2\lambda }{\tan \left( x-3 \right)}\]
We know that \[\displaystyle \lim_{x \to {{3}^{+}}}\dfrac{\tan \left( x-3 \right)}{\left( x-3 \right)}=1\], therefore we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\dfrac{2\lambda }{4}\]
\[\Rightarrow \dfrac{\lambda }{2}\]
Therefore, right-hand side limit of the function is equal to \[\dfrac{\lambda }{2}\]which means that
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}=\dfrac{\lambda }{2}\]
We know that \[\displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)\], therefore we can write:
\[\Rightarrow -\infty =\dfrac{\lambda }{2}\]
\[\Rightarrow \lambda =-\infty \]
As we know the value of \[\lambda \], we can now find the value of \[8100\lambda \], we get:
\[\Rightarrow 8100\lambda =8100\left( -\infty \right)\]
\[\Rightarrow 8100\lambda =-\infty \]
Therefore, the value of \[8100\lambda \] is equal to \[-\infty \].
Note: Students can make calculation mistakes, while solving this type of question. It’s better to recheck the answer after solving it. Also remember the key points, that are \[\displaystyle \lim_{x \to b}\dfrac{{{a}^{x}}-1}{x}=1\], \[\left( a \right)\ln \left( b \right)=\ln {{\left( b \right)}^{a}}\], \[{{e}^{\ln a}}=a\] and \[1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}\].
& \dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left[ x+1 \right]}}-9}{{{3}^{x}}-27}\text{ ; }x<3 \\
& \lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}\text{ ; }x>3 \\
\end{align} \right.\] is continuous at \[x=3\]. Apply the one of the results of continuous function that is “if a function \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\] “to find the value of \[\lambda \].
Complete step by step answer:
According to the question, given function in question is as follows:
\[f\left( x \right)=\left\{ \begin{align}
& \dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left[ x+1 \right]}}-9}{{{3}^{x}}-27}\text{ ; }x<3 \\
& \lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}\text{ ; }x>3 \\
\end{align} \right.\]
We know that this function is continuous at \[x=3\]. Therefore, according to the definition of continuous function that is if a function \[f\left( x \right)\] is continuous at \[x=a\] then \[\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\] .
This implies that as per the given in the question, \[f\left( x \right)\] is continuous at \[x=3\] then \[\displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)\]
Now, let’s first find the limit of the left-hand side of the function.
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( {{e}^{\left( \left( x+3 \right)\ln 27 \right)}} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
Apply the property of logarithm that is \[\left( a \right)\ln \left( b \right)=\ln {{\left( b \right)}^{a}}\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( {{e}^{\left( \ln {{27}^{\left( x+3 \right)}} \right)}} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
We know that \[{{e}^{\ln a}}=a\], therefore we can rewrite above expression as follows:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( {{27}^{\left( x+3 \right)}} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( 27 \right)}^{\dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)}}-9}{{{3}^{x}}-27}\]
Now take the term \[9\] common from the numerator of the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{9.\left( {{27}^{\dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3}}}-1 \right)}{27\left( \dfrac{{{3}^{x}}}{27}-1 \right)}\]
Now take the term \[27\] common from the denominator of the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{9.\left( {{27}^{\dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3}}}-1 \right)}{27\left( {{3}^{x-3}}-1 \right)}\]
After this multiply and divide term \[\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( {{27}^{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}}-1 \right)}{3\left( {{3}^{x-3}}-1 \right)}\times \dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}\]
Apply the result that \[\displaystyle \lim_{x \to b}\dfrac{{{a}^{x}}-1}{x}=1\], therefore we can say that \[\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( {{27}^{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}}-1 \right)}{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}=1\], hence, we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{3\left( {{3}^{x-3}}-1 \right)}\]
After this multiply and divide term \[\left( x-3 \right)\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{3\left( {{3}^{\left( x-3 \right)}}-1 \right)\times \dfrac{\left( x-3 \right)}{\left( x-3 \right)}}\]
Apply the result that \[\displaystyle \lim_{x \to b}\dfrac{{{a}^{x}}-1}{x}=1\], therefore we can say that \[\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( {{3}^{\left( x-3 \right)}}-1 \right)}{\left( x-3 \right)}=1\], hence, we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{1}{27}\left( x+3 \right)\left( x+1 \right)-\dfrac{2}{3} \right)}{3\left( x-3 \right)}\]
Now, take the LCM of the terms in the numerator, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( \dfrac{\left( x+3 \right)\left( x+1 \right)-18}{27} \right)}{3\left( x-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{\left( x+3 \right)\left( x+1 \right)-18}{3\times 27\times \left( x-3 \right)}\]
After simplifying this further we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{x}^{2}}+x+3x+3-18}{3\times 27\times \left( x-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{x}^{2}}+4x-15}{81\left( x-3 \right)}\]
Now, apply the left-hand side limit that is putting \[x=3-h\] in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{{{\left( 3-h \right)}^{2}}+4\left( 3-h \right)-15}{81\left( 3-h-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{{{\left( 3-h \right)}^{2}}+4\left( 3-h \right)-15}{81\left( -h \right)}\]
\[\Rightarrow \dfrac{{{\left( 3-0 \right)}^{2}}+4\left( 3-0 \right)-15}{81\left( -0 \right)}\]
\[\Rightarrow -\infty \]
Therefore, left-hand side limit of the function is equal to \[-\infty \]which means that\[\Rightarrow \displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{-}}}\dfrac{{{\left( \exp \left\{ \left( x+3 \right)\ln 27 \right\} \right)}^{\dfrac{1}{27}\left( x+1 \right)}}-9}{{{3}^{x}}-27}=-\infty \]
Now find the right-hand side limit of the function.
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}\]
Apply the identity that is \[1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}\]in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{2{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\left( x-3 \right)\tan \left( x-3 \right)}\]
Now multiply and divide \[\dfrac{\left( x-3 \right)}{4}\]in the above expression, we get:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{\dfrac{\left( x-3 \right)}{4}}{\dfrac{\left( x-3 \right)}{4}}\dfrac{2{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\left( x-3 \right)\tan \left( x-3 \right)}\]
\[\Rightarrow \displaystyle \lim_{x \to +3}\lambda .\dfrac{\left( x-3 \right)}{4}.\dfrac{2{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\dfrac{{{\left( x-3 \right)}^{2}}}{4}\tan \left( x-3 \right)}\]
Apply the result that \[\displaystyle \lim_{x \to b}\dfrac{{{\sin }^{2}}x}{{{x}^{2}}}=1\], therefore we can say that \[\displaystyle \lim_{x \to {{3}^{+}}}\dfrac{{{\sin }^{2}}\dfrac{\left( x-3 \right)}{2}}{\dfrac{{{\left( x-3 \right)}^{2}}}{4}}=1\], hence, we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\dfrac{\left( x-3 \right)}{4}.\dfrac{2\lambda }{\tan \left( x-3 \right)}\]
We know that \[\displaystyle \lim_{x \to {{3}^{+}}}\dfrac{\tan \left( x-3 \right)}{\left( x-3 \right)}=1\], therefore we can write above expression as:
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}\dfrac{2\lambda }{4}\]
\[\Rightarrow \dfrac{\lambda }{2}\]
Therefore, right-hand side limit of the function is equal to \[\dfrac{\lambda }{2}\]which means that
\[\Rightarrow \displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}\lambda .\dfrac{1-\cos \left( x-3 \right)}{\left( x-3 \right)\tan \left( x-3 \right)}=\dfrac{\lambda }{2}\]
We know that \[\displaystyle \lim_{x \to {{3}^{-}}}f\left( x \right)=\displaystyle \lim_{x \to {{3}^{+}}}f\left( x \right)\], therefore we can write:
\[\Rightarrow -\infty =\dfrac{\lambda }{2}\]
\[\Rightarrow \lambda =-\infty \]
As we know the value of \[\lambda \], we can now find the value of \[8100\lambda \], we get:
\[\Rightarrow 8100\lambda =8100\left( -\infty \right)\]
\[\Rightarrow 8100\lambda =-\infty \]
Therefore, the value of \[8100\lambda \] is equal to \[-\infty \].
Note: Students can make calculation mistakes, while solving this type of question. It’s better to recheck the answer after solving it. Also remember the key points, that are \[\displaystyle \lim_{x \to b}\dfrac{{{a}^{x}}-1}{x}=1\], \[\left( a \right)\ln \left( b \right)=\ln {{\left( b \right)}^{a}}\], \[{{e}^{\ln a}}=a\] and \[1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}\].
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