Question

#Quality Question

# Limits

Find the value of
​

Answer 1

Answer:

$\huge{ \pink{36}}$

Step-by-step explanation:

$\displaystyle{\lim} \: x \longrightarrow2 \: \: \: \dfrac{ {3}^{x} + {3}^{3 - x} - 12 }{ {3}^{ - \frac{x}{2} } - {3}^{1 - x} } \\ = \lim \: x \longrightarrow2 \: \: \: \dfrac{ {3}^{x} + {3}^{3} {3}^{ - x} - 12 }{ ({ {3}^{ - x}) }^{ \frac{1}{2} } - 3 \: {3}^{ - x} } \\ let \: {3}^{x} = y \\when \: \: x \longrightarrow2 \\ \: \: y \longrightarrow9 \\ = \lim \: y \longrightarrow9 \: \: \: \dfrac{y + \dfrac{27}{y} - 12}{ {( \dfrac{1}{y} )}^{ \dfrac{1}{2} } - \dfrac{3}{y} } \\ = \lim \: y \longrightarrow \: 9 \: \dfrac{ \dfrac{ {y}^{2} + 27 - 12y}{y} }{ \dfrac{ \sqrt{y} - 3 }{y} } \\ = \lim \: y \longrightarrow 9 \: \: (\dfrac{ {y}^{2} - 12y + 27 }{ \sqrt{y} - 3 } ) \\ = \lim \: y \longrightarrow9 \: \: \dfrac{ {y}^{2} - 9y - 3y + 27}{ \sqrt{y} - 3} \\ = \lim \: y \longrightarrow9 \: \: \: \dfrac{(y - 9) \: (y - 3)}{ \sqrt{y} - 3 } \\ = \lim \: y \longrightarrow9 \: \: \dfrac{ ({( \sqrt{y}) }^{2} - {(3)}^{2} ) \: \: (y - 3) }{( \sqrt{y} - 3) } \\ = \lim \: y \longrightarrow9 \: \: \dfrac{( \sqrt{y} + 3) \: ( \sqrt{y} - 3) \: (y - 3)}{ \sqrt{y} - 3} \\ = \lim \: y \longrightarrow9 \: \: ( \sqrt{y} + 3) \: (y - 3) \\ = ( \sqrt{9} + 3) \: (9 - 3) \\ = (3 + 3) \: (6) \\ = 36$

Answer 2

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2}\bf \dfrac{ {3}^{x} + {3}^{3 - x} - 12}{ {3}^{ - \frac{x}{2} } - {3}^{1 - x} }$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{x \to 2}\sf \dfrac{ {3}^{x} + {3}^{3} {3}^{ - x} - 12}{ {3}^{ - \frac{x}{2} } - {3}^{1} {3}^{ - x} }$

$\rm \: = \:\displaystyle\lim_{x \to 2}\sf \dfrac{ {3}^{x} + \dfrac{27}{ {3}^{x} } - 12}{ \dfrac{1}{ {3}^{ \frac{x}{2} } } - \dfrac{3}{ {3}^{x} } }$

$\red{\rm :\longmapsto\:Put \: \bigg(3 \bigg)^{\dfrac{x}{2} } = y}$

So,

$\red{\rm :\longmapsto\: {3}^{x} = {y}^{2}}$

As,

$\red{\rm :\longmapsto\:x \: \to \: 2 \: \: so \: \: y \: \to \: 3}$

So, above limit reduces to

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ {y}^{2} + \dfrac{27}{ {y}^{2} } - 12 }{\dfrac{1}{y} - \dfrac{3}{ {y}^{2} } }$

On taking LCM, we get

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ {y}^{4} + 27 - 12 {y}^{2} }{y - 3}$

If we put directly y = 3, we get indeterminant form,

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ {y}^{4} - 12 {y}^{2} + 27}{y - 3}$

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ {y}^{4} - 9 {y}^{2} - {3y}^{2} + 27}{y - 3}$

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ {y}^{2}( {y}^{2} - 9) - 3( {y}^{2} - 9)}{y - 3}$

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ ({y}^{2} - 3)( {y}^{2} - 9) }{y - 3}$

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ ({y}^{2} - 3)( {y}^{2} - {3}^{2} ) }{y - 3}$

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf \dfrac{ ({y}^{2} - 3)(y + 3)(y - 3) }{y - 3}$

$\rm \: = \: \: \displaystyle\lim_{y \to 3}\sf ({y}^{2} - 3)(y + 3)$

$\rm \: = \: \: (9 - 3)(3 + 3)$

$\rm \: = \: \: 6\times 6$

$\rm \: = \: \: 36$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2}\bf \dfrac{ {3}^{x} + {3}^{3 - x} - 12}{ {3}^{ - \frac{x}{2} } - {3}^{1 - x} } = 36$

Additional Information :-

$\blue{ \boxed{\displaystyle\lim_{x \to 0}\bf \: \frac{sinx}{x} = 1}}$

$\blue{ \boxed{\displaystyle\lim_{x \to 0}\bf \: \frac{tanx}{x} = 1}}$

$\blue{ \boxed{\displaystyle\lim_{x \to 0}\bf \: \frac{log(1 + x)}{x} = 1}}$

$\blue{ \boxed{\displaystyle\lim_{x \to 0}\bf \: \frac{ {e}^{x} - 1}{x} = 1}}$

$\blue{ \boxed{\displaystyle\lim_{x \to 0}\bf \: \frac{ {a}^{x} - 1}{x} = loga}}$

$\blue{ \boxed{\displaystyle\lim_{x \to 0}\bf \: \frac{sin^{ - 1} x}{x} = 1}}$

$\blue{ \boxed{\displaystyle\lim_{x \to 0}\bf \: \frac{tan^{ - 1} x}{x} = 1}}$