Question

If f\left( x \right)=\left\\{ \begin{aligned} & \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{aligned} \right. is continuous at $x=3$, then the value of $8100\lambda$ must be?

Answer 1

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{aligned} & \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{aligned} \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:

& \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{aligned} \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}$$.