Question

plz help me 11th std pcmb Maths problem limits chapter name plz help me ​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \dfrac{ {4}^{x - 1} - {2}^{x} + 1}{ {(x - 1)}^{2} }$

On substituting directly x = 1, we get

$\rm \: = \: \: \dfrac{ {4}^{1 - 1} - {2}^{1} + 1}{ {(1 - 1)}^{2} }$

$\rm \: = \: \: \dfrac{ {4}^{0} - {2}^{1} + 1}{ {(0)}^{2} }$

$\rm \: = \: \: \dfrac{1 - 2 + 1}{0}$

$\rm \: = \: \: \dfrac{0}{0}$

$\rm \: = \: \: which \: is \: meaningless.$

To, solve this now,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \dfrac{ {4}^{x - 1} - {2}^{x} + 1}{ {(x - 1)}^{2} }$

$\red{\rm :\longmapsto\:Put \: x = 1 + h, \: as \: x \to \: 1 \: so \: h \: \to \: 0}$

$\rm \: = \: \: \displaystyle\lim_{h \to 0} \dfrac{ {4}^{1 + h - 1} - {2}^{1 + h} + 1}{ {(1 + h - 1)}^{2} }$

$\rm \: = \: \: \displaystyle\lim_{h \to 0} \dfrac{ {4}^{h} - {2}^{1 + h} + 1}{ {h}^{2} }$

$\rm \: = \: \: \displaystyle\lim_{h \to 0} \dfrac{ {4}^{h} - 2.{2}^{h} + 1}{ {h}^{2} }$

$\rm \: = \: \: \displaystyle\lim_{h \to 0} \dfrac{ {2}^{2h} - 2.{2}^{h} + 1}{ {h}^{2} }$

$\rm \: = \: \: \displaystyle\lim_{h \to 0} \dfrac{ ({2}^{h})^{2} - 2.{2}^{h} + 1}{ {h}^{2} }$

We know,

$\underbrace{ \boxed{ \bf \: {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2}}}$

$\rm \: = \: \: \displaystyle\lim_{h \to 0} \dfrac{ ({2}^{h} - 1)^{2}}{ {h}^{2} }$

$\rm \: = \: \: \displaystyle\lim_{h \to 0} \dfrac{ ({2}^{h} - 1)}{ {h}} \times \displaystyle\lim_{h \to 0} \dfrac{ ({2}^{h} - 1)}{ {h}}$

We know,

$\underbrace{ \boxed{ \bf \: \displaystyle\lim_{x \to 0} \dfrac{ ({a}^{x} - 1)}{ {x}} = loga}}$

So, using this formula, we have .

$\rm \: = \: \: log2 \times log2$

$\rm \: = \: \: {(log2)}^{2}$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \bf \: \dfrac{ {4}^{x - 1} - {2}^{x} + 1}{ {(x - 1)}^{2} } = {(log2)}^{2}$

Alternative Method :-

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \dfrac{ {4}^{x - 1} - {2}^{x} + 1}{ {(x - 1)}^{2} }$

Using L - Hospital Rule, we get

$\rm \: = \: \:\displaystyle\lim_{x \to 1} \dfrac{ {4}^{x - 1}log4 - {2}^{x}log2}{2 {(x - 1)}}$

$\rm \: = \: \:\displaystyle\lim_{x \to 1} \dfrac{ {4}^{x - 1}(log4)^{2} - {2}^{x}(log2)^{2} }{2}$

$\rm \: = \: \: \dfrac{ {4}^{1 - 1}(log4)^{2} - {2}^{1}(log2)^{2} }{2}$

$\rm \: = \: \: \dfrac{ {4}^{0}(log4)^{2} - {2}(log2)^{2} }{2}$

$\rm \: = \: \: \dfrac{(2log2)^{2} - {2}(log2)^{2} }{2}$

$\rm \: = \: \: \dfrac{4(log2)^{2} - {2}(log2)^{2} }{2}$

$\rm \: = \: \: \dfrac{(log2)^{2}}{2}$

$\rm \: = \: \: {(log2)}^{2}$

Hence,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1} \bf \: \dfrac{ {4}^{x - 1} - {2}^{x} + 1}{ {(x - 1)}^{2} } = {(log2)}^{2}$

Formula Used :-

$\rm :\longmapsto\:\dfrac{d}{dx} {a}^{x} = {a}^{x}loga$

$\rm :\longmapsto\:\dfrac{d}{dx}k = 0$

$\rm :\longmapsto\:\dfrac{d}{dx}x = 1$

$\rm :\longmapsto\:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\rm :\longmapsto\: log( {x}^{y} ) = y \: logx$