Question

lim tends 0 log(1+ax) - log (1-bx) ÷ x = 8 , Given a - b = 4 .Then a÷b =​

Answer 1

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

Given that,

$\rm :\longmapsto\:a - b = 4 - - - (1)$

and

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{log(1 + ax) - log(1 - bx)}{x} = 8$

can be rewritten as

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{log(1 + ax)}{x} - \displaystyle\lim_{x \to 0}\rm \frac{log(1 - bx)}{x} = 8$

can be further rewritten as

$\rm :\longmapsto\:a\displaystyle\lim_{x \to 0}\rm \frac{log(1 + ax)}{ax} + b \displaystyle\lim_{x \to 0}\rm \frac{log(1 - bx)}{ - bx} = 8$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}}$

So, using this identity, we get

$\rm :\longmapsto\:a + b = 8 - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2a = 12$

$\rm\implies \:\boxed{\tt{ \: \: a \: = \: 6 \: \: }}$

On Subtracting equation (1) from equation (2), we get

$\rm :\longmapsto\:2b = 4$

$\rm\implies \:\boxed{\tt{ \: \: b \: = \: 2 \: \: }}$

Hence,

$\bf\implies \:\dfrac{a}{b} = \dfrac{6}{2} = 3$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}}$

Answer 2

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

Given that,

$\rm :\longmapsto\:a - b = 4 - - - (1)$

and

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{log(1 + ax) - log(1 - bx)}{x} = 8$

can be rewritten as

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{log(1 + ax)}{x} - \displaystyle\lim_{x \to 0}\rm \frac{log(1 - bx)}{x} = 8$

can be further rewritten as

$\rm :\longmapsto\:a\displaystyle\lim_{x \to 0}\rm \frac{log(1 + ax)}{ax} + b \displaystyle\lim_{x \to 0}\rm \frac{log(1 - bx)}{ - bx} = 8$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}}$

So, using this identity, we get

$\rm :\longmapsto\:a + b = 8 - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2a = 12$

$\rm\implies \:\boxed{\tt{ \: \: a \: = \: 6 \: \: }}$

On Subtracting equation (1) from equation (2), we get

$\rm :\longmapsto\:2b = 4$

$\rm\implies \:\boxed{\tt{ \: \: b \: = \: 2 \: \: }}$

Hence,

$\bf\implies \:\dfrac{a}{b} = \dfrac{6}{2} = 3$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}}$

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}}$