Q: Show that \$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}}\$ does not exist.

Given\$\displaystyle \lim_{x\rightarrow 0} \dfrac{e^{\dfrac{1}{x}}-1}{e^{\dfrac{1}{x}}+1}\$Left hand limit.L.H.L \$ = \displaystyle \lim_{x\rightarrow 0^{-}} \dfrac{e^{\dfrac{1}{0-}}-1}{e^{\dfrac{1}{0-}}+1}= \dfrac{e^{-\infty }-1}{e^{-\infty }+1}= \dfrac{0-1}{0+1} = -1\$we know that \$ e^{\infty } = \infty \$\$\displaystyle e^{-\infty } = \dfrac{1}{e^{\infty }} = \dfrac{1}{\infty } \Rightarrow 0\$Right hand limit R.H.L = \$ \displaystyle \lim_{x\rightarrow 0^{+}}\dfrac{{\dfrac{1}{e^{0+}}}-1}{e^{\dfrac{1}{0+}}+1} = \dfrac{e^{\infty }-1}{e^{\infty }+1} = \dfrac{e^{\infty }(1-\dfrac{1}{e^{\infty }})}{e^{\infty }(1+\dfrac{1}{e^{\infty }})}\$\$ = \dfrac{1-0}{1+0} = 1\$So, here \$ L.H.L \neq R.H.L\$So, limit does not exist.

Q: \$\lim_{x\to1}(1-x)\tan \left (\frac{\pi x}{2}\right)\$ is equal to -

Let \$t=1-x\$\$\Rightarrow x=1-t\$\$x \rightarrow 1\$\$\Rightarrow t=0\$Now \$\lim_{t \to 0} t\tan \left (\frac{\pi}{2}(1-t) \right )\$\$=\lim_{t \to 0} t\tan \left (\frac{\pi}{2}-\frac{\pi}{2}t \right )\$\$=\lim_{t \to 0} t\cot \left (\frac{\pi}{2}-t \right )\$\$=\lim_{t \to 0} \frac{t}{\tan \frac{\pi}{2}t}\$Now multiply & divide the above expression by \$\frac{\pi}{2}\$ we get\$=\lim_{t \to 0}\frac{2}{\pi} \left ( \frac{\frac{\pi}{2}t}{\tan \frac{\pi}{2}t} \right )=\frac{2}{\pi}\$

Q: \$\lim_{x\rightarrow 0}(\frac{1}{x^{2}}-\frac{1}{sin^{2}x})\$

Consider the given function: \$ f(x)=\underset{x\to 0}{\mathop{\lim }}\,(\dfrac{1}{{{x}^{2}}}-\dfrac{1}{{{\sin }^{2}}x}) \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,(\dfrac{{{\sin }^{2}}x-{{x}^{2}}}{{{x}^{2}}{{\sin }^{2}}x}).\dfrac{{{x}^{2}}}{{{x}^{2}}} \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,(\dfrac{{{\sin }^{2}}x-{{x}^{2}}}{{{x}^{4}}})(\dfrac{{{x}^{2}}}{{{\sin }^{2}}x}) \$ \$ we\,\,know\,\,that\,\,(\dfrac{0}{0}\,)\,\,form, \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2\sin x\cos x-2x}{4{{x}^{3}}}.1 \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2(-\sin x\sin x+\cos x\cos x)-2}{12{{x}^{2}}} \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2({{\cos }^{2}}x-{{\sin }^{2}}x)-2}{12{{x}^{2}}} \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2\cos 2x-2}{12{{x}^{2}}} \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{-4\sin 2x-0}{24{{x}^{2}}} \$ \$ =\underset{x\to 0}{\mathop{\lim }}\,\,-4(\dfrac{\sin 2x}{2x}).(\dfrac{1}{2}) \$ \$ =\dfrac{-4}{12}.1\,\,\,=\,\,\,-\dfrac{1}{3} \$ Hence this is the answer.

Q: the value of \$\displaystyle x\xrightarrow { lim } 0\,\frac{\sin{x}^{0}}{x}\$

We have \$\lim _{ x\rightarrow 0 }{ \cfrac { \sin { x } }{ x } } =1\$ when \$x\$ is in radiansalso, \${ x }^{ \circ }=\cfrac { \pi }{ 180 } x\$ radians\$\therefore \lim _{ x\rightarrow 0 }{ \cfrac { \sin { { x }^{ \circ } } }{ x } = } \lim _{ x\rightarrow 0 }{ \cfrac { \sin { \cfrac { \pi x }{ 180 } } }{ x } } =\lim _{ x\rightarrow 0 }{ \cfrac { \sin { (\cfrac { \pi x }{ 180 } ) } }{ (\cfrac { \pi x }{ 180 } )\times (\cfrac { 180 }{ \pi } ) } } \\ \Rightarrow \lim _{ x\rightarrow 0 }{ \cfrac { \sin { { x }^{ \circ } } }{ x } } =\cfrac { \pi }{ 180 } \lim _{ x\rightarrow 0 }{ \cfrac { \sin { (\cfrac { \pi x }{ 180 } ) } }{ (\cfrac { \pi x }{ 180 } ) } } =\cfrac { \pi }{ 180 } \$

Q: Let \$\alpha\$ and \$\beta\$ be the distinct roots of \$ax^{2}+bx+c=0\$ then \$\displaystyle \lim_{x\rightarrow a}\$ \$\displaystyle \frac{1-\cos(ax^{2}+bx+c)}{(x-\alpha)^{2}}\$ is equal to-

As \$\alpha\$ and \$\beta\$ are roots of \$a{x}^{2}+bx+c=0\$, thensum of roots \$\displaystyle\alpha+\beta=-\frac{b}{a}\$and product of roots \$\displaystyle\alpha\beta=\frac{c}{a}\$Given limit = \$\displaystyle\lim_{x\rightarrow a}\$ \$\displaystyle \frac{1-cosa(x-\alpha )(x-\beta )}{(x-\alpha )^{2}}\$ \$\displaystyle=\lim_{x\rightarrow a}\displaystyle\frac{2sin^{2}\left ( a\displaystyle\frac{(x-a)(x-\beta )}{2} \right )}{(x-\alpha )^{2}}\$ \$=\displaystyle\lim_{x\rightarrow \alpha }\frac{2}{(x-\alpha )^{2}}\times\frac{sin^{2}(a)\displaystyle\frac{(x-\alpha )(x-\beta )}{2}}{\displaystyle\frac{a^{2}(x-\alpha )^{2}(x-\beta )^{2}}{4}}\times\frac{a^{2}(x-\alpha )^{2}(x-\beta )^{2}}{4}=\frac{a^{2}(\alpha -\beta )^{2}}{2}\$

Q: Evaluate : \$\lim _ { x \rightarrow 0 } \dfrac { ( 1 + x ) ^ { 1 / x } - e } { x } =?\$

\$\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 1+x \right) }^{ 1/x }-e }{ x } \$\$=\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 1+x \right) }^{ 1/x }\left[ \dfrac { x }{ 1+x } -ln\left( x+1 \right) \right] /{ x }^{ 2 } }{ 1 } \$\$=\underset { x\rightarrow 0 }{ lt } \dfrac { e\left( \dfrac { \left( 1+x \right) -x }{ { \left( 1+x \right) }^{ 2 } } -\dfrac { 1 }{ \left( x+1 \right) } \right) }{ 2x } \$\$=e\underset { x\rightarrow 0 }{ lt } \dfrac { 1-\left( x+1 \right) }{ { \left( x+1 \right) }^{ 2 }\times 2x } \$\$e\underset { x\rightarrow 0 }{ lt } \dfrac { x }{ { \left( x+1 \right) }^{ 2 }2x } \$\$=-e\underset { x\rightarrow 0 }{ lt } \dfrac { x }{ { \left( x+1 \right) }^{ 2 }\times 2x } \$\$=\dfrac { -e }{ 2 } \$

Q: If \$f\left( x \right) =\log _{ x }{ \left( \log { x } \right) } \$, then \$f'\left( x \right)\$ at \$x=e\$ is

Given \$f(x)=\log (\log {x})\$Differentiate it with respect to \$x\$\$f'(x)=\dfrac{1}{\log{x}}.\dfrac{1}{x}\$Substituting \$x=e\$\$f'(x)=\dfrac{1}{\log{e}}.\dfrac{1}{e}\$\$f'(x)=\dfrac{1}{e}\$

Q: \$\lim_{x \rightarrow 1}\dfrac {1-x^{-1/3}}{1-x^{-2/3}}\$

\$ \displaystyle \lim_{x\rightarrow 1} \frac{1-x^{-1/3}}{1-x^{-2/3}} \$Here putting the limit \$ x\rightarrow 1 \$ we get \$ \dfrac{0}{0} \$ from (undefined form )None we can apply L-Hospital Rule \$ \displaystyle\Rightarrow \lim_{x\rightarrow 1} \dfrac{d(1-x^{1-/3})}{d(1-x^{-2/3})} \$\$ \displaystyle \Rightarrow \lim_{x\rightarrow 1} \dfrac{0-(\dfrac{-1}{3}x^{-1/3-1})}{0-(\dfrac{-2}{3}x^{-2/3}-1)} \$ \$\{\dfrac{d}{dx}(x^{n}) = nx^{n-1} = d(x)^{n} \$\$ \displaystyle\Rightarrow \lim_{x\rightarrow 1} \dfrac{+\dfrac{1}{3}x^{-4/3}}{+\dfrac{2}{3}x^{-5/3}} \$\$ \displaystyle\Rightarrow \lim_{x\rightarrow 1} \dfrac{1}{2}x^{-4/3}+\dfrac{5}{3} \$\$ \displaystyle\Rightarrow \lim_{x\rightarrow 1} \dfrac{1}{2}x\dfrac{1}{3} \$\$ \displaystyle \Rightarrow \dfrac{1}{2}(1)^{1/3}\Rightarrow \dfrac{1}{2} \$

Q: Evaluate\$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2 - \sqrt {1 + \cos x} }}{{{{\sin }^2}x}}\$

Solution -\$ \lim_{x\rightarrow 0} \frac{\sqrt{2}-\sqrt{1+cos\,x}}{sin^{2}\,x} \$Apply L - Hospital Rule -\$\lim_{x\rightarrow 0}\frac{-1(-sin\,x)}{2\frac{\sqrt{1+cos\,x}}{2\,sin\,x\,cos\,x}} \$ \$ \lim_{x\rightarrow 0} \frac{1}{4\,cos\,x\sqrt{1+cos\,x}} \$\$ = \frac{1}{4\sqrt{2}} \$

Q: Evaluate: \$\displaystyle \lim_{x \rightarrow 0}{\left( \frac{a^x + b^x + c^x}{3} \right)^{2/x}}\$

\$\displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cfrac { { a }^{ x }+{ b }^{ x }+{ c }^{ x } }{ 3 } \right) }^{ { 2 }/{ x } } } ={ \left( \cfrac { 3 }{ 3 } \right) }^{ { 2 }/{ 0 } }\$\$={ 1 }^{ \infty }\$form\$={ e }^{ \displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cfrac { { a }^{ x }+{ b }^{ x }+{ c }^{ x } }{ 3 } -1 \right) }^{ { 2 }/{ x } } } }\$\$={ e }^{ \displaystyle \lim _{ x\rightarrow 0 }{ { \cfrac { 2 }{ 3 } \left( \cfrac { { a }^{ x }-1 }{ x } +\cfrac { { b }^{ x }-1 }{ x } +\cfrac { { c }^{ x }-1 }{ x } \right) } } }\$\$={ e }^{ \cfrac { 2 }{ 3 } \left( \log { a } +\log { b } +\log { c } \right) }\$\$={ e }^{ \log _{ e }{ { \left( abc \right) }^{ { 2 }/{ 3 } } } }\$\$={ (abc) }^{ { 2 }/{ 3 } }\$

Question 1

Show that $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}}$ does not exist.

Accepted Answer

Given$\displaystyle  \lim_{x\rightarrow 0} \dfrac{e^{\dfrac{1}{x}}-1}{e^{\dfrac{1}{x}}+1}$Left hand limit.L.H.L $ = \displaystyle \lim_{x\rightarrow 0^{-}} \dfrac{e^{\dfrac{1}{0-}}-1}{e^{\dfrac{1}{0-}}+1}= \dfrac{e^{-\infty }-1}{e^{-\infty }+1}= \dfrac{0-1}{0+1} = -1$we know that $ e^{\infty } = \infty $$\displaystyle  e^{-\infty } = \dfrac{1}{e^{\infty }} = \dfrac{1}{\infty } \Rightarrow 0$Right hand limit R.H.L = $ \displaystyle \lim_{x\rightarrow 0^{+}}\dfrac{{\dfrac{1}{e^{0+}}}-1}{e^{\dfrac{1}{0+}}+1} = \dfrac{e^{\infty }-1}{e^{\infty }+1} = \dfrac{e^{\infty }(1-\dfrac{1}{e^{\infty }})}{e^{\infty }(1+\dfrac{1}{e^{\infty }})}$$ = \dfrac{1-0}{1+0} = 1$So, here $ L.H.L \neq R.H.L$So, limit does not exist.

Question 2

$\lim_{x\to1}(1-x)\tan \left (\frac{\pi x}{2}\right)$ is equal to -

Accepted Answer

Let $t=1-x$$\Rightarrow x=1-t$$x \rightarrow 1$$\Rightarrow t=0$Now $\lim_{t \to 0} t\tan \left (\frac{\pi}{2}(1-t) \right )$$=\lim_{t \to 0} t\tan \left (\frac{\pi}{2}-\frac{\pi}{2}t \right )$$=\lim_{t \to 0} t\cot \left (\frac{\pi}{2}-t \right )$$=\lim_{t \to 0} \frac{t}{\tan \frac{\pi}{2}t}$Now multiply & divide the above expression by $\frac{\pi}{2}$ we get$=\lim_{t \to 0}\frac{2}{\pi} \left ( \frac{\frac{\pi}{2}t}{\tan \frac{\pi}{2}t} \right )=\frac{2}{\pi}$

Question 3

$\lim_{x\rightarrow 0}(\frac{1}{x^{2}}-\frac{1}{sin^{2}x})$

Accepted Answer

Consider the&#10;given function:&#10;&#10;  $ f(x)=\underset{x\to 0}{\mathop{\lim }}\,(\dfrac{1}{{{x}^{2}}}-\dfrac{1}{{{\sin&#10;}^{2}}x}) $ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,(\dfrac{{{\sin&#10;}^{2}}x-{{x}^{2}}}{{{x}^{2}}{{\sin }^{2}}x}).\dfrac{{{x}^{2}}}{{{x}^{2}}} $ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,(\dfrac{{{\sin&#10;}^{2}}x-{{x}^{2}}}{{{x}^{4}}})(\dfrac{{{x}^{2}}}{{{\sin }^{2}}x}) $ &#10;&#10; $ we\,\,know\,\,that\,\,(\dfrac{0}{0}\,)\,\,form,&#10;$ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2\sin&#10;x\cos x-2x}{4{{x}^{3}}}.1 $ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2(-\sin&#10;x\sin x+\cos x\cos x)-2}{12{{x}^{2}}} $ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2({{\cos&#10;}^{2}}x-{{\sin }^{2}}x)-2}{12{{x}^{2}}} $ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{2\cos&#10;2x-2}{12{{x}^{2}}} $ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,\,\dfrac{-4\sin&#10;2x-0}{24{{x}^{2}}} $ &#10;&#10; $ =\underset{x\to 0}{\mathop{\lim }}\,\,-4(\dfrac{\sin&#10;2x}{2x}).(\dfrac{1}{2}) $ &#10;&#10; $ =\dfrac{-4}{12}.1\,\,\,=\,\,\,-\dfrac{1}{3}&#10;$&#10;&#10;Hence&#10;this is the answer.

Question 4

the value of $\displaystyle x\xrightarrow { lim } 0\,\frac{\sin{x}^{0}}{x}$

Accepted Answer

We have $\lim _{ x\rightarrow 0 }{ \cfrac { \sin { x }  }{ x }  } =1$ when $x$ is in radiansalso, ${ x }^{ \circ  }=\cfrac { \pi  }{ 180 } x$ radians$\therefore \lim _{ x\rightarrow 0 }{ \cfrac { \sin { { x }^{ \circ  } }  }{ x } = } \lim _{ x\rightarrow 0 }{ \cfrac { \sin { \cfrac { \pi x }{ 180 }  }  }{ x }  } =\lim _{ x\rightarrow 0 }{ \cfrac { \sin { (\cfrac { \pi x }{ 180 } ) }  }{ (\cfrac { \pi x }{ 180 } )\times (\cfrac { 180 }{ \pi  } ) }  } \ \Rightarrow \lim _{ x\rightarrow 0 }{ \cfrac { \sin { { x }^{ \circ  } }  }{ x }  } =\cfrac { \pi  }{ 180 } \lim _{ x\rightarrow 0 }{ \cfrac { \sin { (\cfrac { \pi x }{ 180 } ) }  }{ (\cfrac { \pi x }{ 180 } ) }  } =\cfrac { \pi  }{ 180 } $

Question 5

Let $\alpha$ and $\beta$ be the distinct roots of $ax^{2}+bx+c=0$  then $\displaystyle \lim_{x\rightarrow a}$ $\displaystyle \frac{1-\cos(ax^{2}+bx+c)}{(x-\alpha)^{2}}$ is equal to-

Accepted Answer

As $\alpha$ and $\beta$ are roots of $a{x}^{2}+bx+c=0$, thensum of roots $\displaystyle\alpha+\beta=-\frac{b}{a}$and product of roots $\displaystyle\alpha\beta=\frac{c}{a}$Given limit = $\displaystyle\lim_{x\rightarrow a}$ $\displaystyle \frac{1-cosa(x-\alpha )(x-\beta )}{(x-\alpha )^{2}}$ $\displaystyle=\lim_{x\rightarrow a}\displaystyle\frac{2sin^{2}\left ( a\displaystyle\frac{(x-a)(x-\beta )}{2} \right )}{(x-\alpha )^{2}}$  $=\displaystyle\lim_{x\rightarrow \alpha }\frac{2}{(x-\alpha )^{2}}\times\frac{sin^{2}(a)\displaystyle\frac{(x-\alpha )(x-\beta )}{2}}{\displaystyle\frac{a^{2}(x-\alpha )^{2}(x-\beta )^{2}}{4}}\times\frac{a^{2}(x-\alpha )^{2}(x-\beta )^{2}}{4}=\frac{a^{2}(\alpha -\beta )^{2}}{2}$

Question 6

Evaluate : $\lim _ { x \rightarrow 0 } \dfrac { ( 1 + x ) ^ { 1 / x } - e } { x } =?$

Accepted Answer

$\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 1+x \right)  }^{ 1/x }-e }{ x } $$=\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 1+x \right)  }^{ 1/x }\left[ \dfrac { x }{ 1+x } -ln\left( x+1 \right)  \right] /{ x }^{ 2 } }{ 1 } $$=\underset { x\rightarrow 0 }{ lt } \dfrac { e\left( \dfrac { \left( 1+x \right) -x }{ { \left( 1+x \right)  }^{ 2 } } -\dfrac { 1 }{ \left( x+1 \right)  }  \right)  }{ 2x } $$=e\underset { x\rightarrow 0 }{ lt } \dfrac { 1-\left( x+1 \right)  }{ { \left( x+1 \right)  }^{ 2 }\times 2x } $$e\underset { x\rightarrow 0 }{ lt } \dfrac { x }{ { \left( x+1 \right)  }^{ 2 }2x } $$=-e\underset { x\rightarrow 0 }{ lt } \dfrac { x }{ { \left( x+1 \right)  }^{ 2 }\times 2x } $$=\dfrac { -e }{ 2 } $

Question 7

If $f\left( x \right) =\log _{ x }{ \left( \log { x }  \right)  } $, then $f'\left( x \right)$ at $x=e$ is

Accepted Answer

Given $f(x)=\log (\log {x})$Differentiate it with respect to $x$$f'(x)=\dfrac{1}{\log{x}}.\dfrac{1}{x}$Substituting $x=e$$f'(x)=\dfrac{1}{\log{e}}.\dfrac{1}{e}$$f'(x)=\dfrac{1}{e}$

Question 8

$\lim_{x \rightarrow 1}\dfrac {1-x^{-1/3}}{1-x^{-2/3}}$

Accepted Answer

$ \displaystyle \lim_{x\rightarrow 1} \frac{1-x^{-1/3}}{1-x^{-2/3}} $Here putting the limit $ x\rightarrow 1 $ we get $ \dfrac{0}{0} $ from (undefined form )None we can apply L-Hospital Rule $ \displaystyle\Rightarrow \lim_{x\rightarrow 1} \dfrac{d(1-x^{1-/3})}{d(1-x^{-2/3})} $$ \displaystyle \Rightarrow \lim_{x\rightarrow 1} \dfrac{0-(\dfrac{-1}{3}x^{-1/3-1})}{0-(\dfrac{-2}{3}x^{-2/3}-1)} $     $\{\dfrac{d}{dx}(x^{n}) = nx^{n-1} = d(x)^{n} $$ \displaystyle\Rightarrow  \lim_{x\rightarrow 1} \dfrac{+\dfrac{1}{3}x^{-4/3}}{+\dfrac{2}{3}x^{-5/3}} $$ \displaystyle\Rightarrow  \lim_{x\rightarrow 1} \dfrac{1}{2}x^{-4/3}+\dfrac{5}{3} $$ \displaystyle\Rightarrow  \lim_{x\rightarrow 1} \dfrac{1}{2}x\dfrac{1}{3} $$ \displaystyle \Rightarrow \dfrac{1}{2}(1)^{1/3}\Rightarrow \dfrac{1}{2} $

Question 9

Evaluate$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2  - \sqrt {1 + \cos x} }}{{{{\sin }^2}x}}$

Accepted Answer

Solution -$ \lim_{x\rightarrow 0} \frac{\sqrt{2}-\sqrt{1+cos\,x}}{sin^{2}\,x} $Apply L - Hospital Rule -$\lim_{x\rightarrow 0}\frac{-1(-sin\,x)}{2\frac{\sqrt{1+cos\,x}}{2\,sin\,x\,cos\,x}} $ $ \lim_{x\rightarrow 0} \frac{1}{4\,cos\,x\sqrt{1+cos\,x}} $$ = \frac{1}{4\sqrt{2}} $

Question 10

Evaluate: $\displaystyle \lim_{x \rightarrow 0}{\left( \frac{a^x + b^x + c^x}{3} \right)^{2/x}}$

Accepted Answer

$\displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cfrac { { a }^{ x }+{ b }^{ x }+{ c }^{ x } }{ 3 }  \right)  }^{ { 2 }/{ x } } } ={ \left( \cfrac { 3 }{ 3 }  \right)  }^{ { 2 }/{ 0 } }$$={ 1 }^{ \infty  }$form$={ e }^{ \displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cfrac { { a }^{ x }+{ b }^{ x }+{ c }^{ x } }{ 3 } -1 \right)  }^{ { 2 }/{ x } } }  }$$={ e }^{ \displaystyle \lim _{ x\rightarrow 0 }{ { \cfrac { 2 }{ 3 } \left( \cfrac { { a }^{ x }-1 }{ x } +\cfrac { { b }^{ x }-1 }{ x } +\cfrac { { c }^{ x }-1 }{ x }  \right)  } }  }$$={ e }^{ \cfrac { 2 }{ 3 } \left( \log { a } +\log { b } +\log { c }  \right)  }$$={ e }^{ \log _{ e }{ { \left( abc \right)  }^{ { 2 }/{ 3 } } }  }$$={ (abc) }^{ { 2 }/{ 3 } }$

Introduction to Limits II

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