Question

the right hand and left hand limit of the function f(x)=(e^1/x+1)÷e^1/x-1)if x=0 and x not eqal to 0​

Answer 1

Question :-

To find the right hand limit and left hand limit of the function

$\rm :\longmapsto\: \dfrac{ {\bigg[e\bigg]}^{ \dfrac{1}{x} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{x} } - 1} \: at \: x = 0$

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

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^{ - } } \frac{ {\bigg[e\bigg]}^{ \dfrac{1}{x} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{x} } - 1}$

To evaluate, such limit, we use method of Substitution

So, Substitute

$\rm :\longmapsto\:x = 0 - h \: \: as \: x \to \: 0 \: \: so \: \: h \: \to \: 0 \:$

So, above can be rewritten as

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{ {\bigg[e\bigg]}^{ \dfrac{1}{ - h} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{ - h} } - 1}$

$\rm \: = \: \dfrac{ {\bigg[e\bigg]}^{ \dfrac{1}{ - 0} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{ - 0} } - 1}$

$\rm \: = \:\dfrac{ {e}^{ - \infty } + 1 }{ {e}^{ - \infty } - 1 }$

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

$\rm \: = \:\dfrac{ 1}{ - 1}$

$\rm \: = \: - 1$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0^{ - } } \frac{ {\bigg[e\bigg]}^{ \dfrac{1}{x} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{x} } - 1} = - 1 \: }}$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0^{ + } } \frac{ {\bigg[e\bigg]}^{ \dfrac{1}{x} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{x} } - 1}$

To evaluate, such limit, we use method of Substitution

So, Substitute

$\rm :\longmapsto\:x = 0 + h = h\: \: as \: x \to \: 0 \: \: so \: \: h \: \to \: 0 \:$

So, above can be rewritten as

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{ {\bigg[e\bigg]}^{ \dfrac{1}{h} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{h} } - 1}$

can be further rewritten as

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{{\bigg[e\bigg]}^{ \dfrac{1}{x} }\bigg[ 1 + {\bigg[e\bigg]}^{ \dfrac{1}{ - h} }\bigg]}{{\bigg[e\bigg]}^{ \dfrac{1}{x} }\bigg[ 1 - {\bigg[e\bigg]}^{ \dfrac{1}{ - h} }\bigg]}$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{1 + {\bigg[e\bigg]}^{ \dfrac{1}{ - h} }}{1 - {\bigg[e\bigg]}^{ \dfrac{1}{ - h} }}$

$\rm \: = \: \dfrac{1 + {\bigg[e\bigg]}^{ \dfrac{1}{ - 0} }}{1 - {\bigg[e\bigg]}^{ \dfrac{1}{ - 0} }}$

$\rm \: = \:\dfrac{1 + {e}^{ - \infty } }{1 - {e}^{ - \infty } }$

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

$\rm \: = \:\dfrac{1}{1}$

$\rm \: = \:1$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0^{ + } } \frac{ {\bigg[e\bigg]}^{ \dfrac{1}{x} } + 1 }{{\bigg[e\bigg]}^{ \dfrac{1}{x} } - 1} = 1 \: }}$