Question

Evaluate the following limits
$\sf (i) lim_{x\to0}\dfrac{sin\;ax}{sin\;bx} \bigg(a,b\neq0\bigg)$
Or,
$\sf (ii)lim_{x\to0}\dfrac{sin\;ax}{bx}$
[Note - If you know any one you may do but better to do both]

Answer 1

HOPE IT HELPS!

Thanks for asking.

Answer 2

$\large\underline{\sf{Solution-(i)}}$

Given function is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{sinax}{sinbx}$

If we substitute directly x = 0, we get

$\rm \:  =  \:\dfrac{sin0}{sin0}$

$\rm \:  =  \:\dfrac{0}{0}$

which is indeterminant form.

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{sinax}{sinbx}$

can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{x \to 0}\dfrac{\dfrac{sinax}{ax} \times ax}{\dfrac{sinbx}{bx} \times bx }$

can further rewritten as

$\rm \:  =  \:\displaystyle\lim_{x \to 0}\dfrac{\dfrac{sinax}{ax} \times a}{\dfrac{sinbx}{bx} \times b}$

$\rm \:  =  \: \dfrac{a}{b} \:\displaystyle\lim_{x \to 0}\dfrac{\dfrac{sinax}{ax}}{\dfrac{sinbx}{bx}}$

$\rm \:  =  \:\dfrac{a}{b} \: \dfrac{\displaystyle\lim_{x \to 0} \frac{sinax}{ax} }{\displaystyle\lim_{x \to 0} \frac{sinbx}{bx} }$

We know,

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

So, using this, we get

$\rm \:  =  \:\dfrac{a}{b} \times \dfrac{1}{1}$

$\rm \:  =  \:\dfrac{a}{b}$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{sinax}{sinbx} = \frac{a}{b} \: }}}$

$\green{\large\underline{\sf{Solution-(ii)}}}$

Given function is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{sinax}{bx}$

If we substitute directly x = 0, we get

$\rm \:  =  \:\dfrac{sin0}{0}$

$\rm \:  =  \:\dfrac{0}{0}$

which is indeterminant form.

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{sinax}{bx}$

can be rewritten as

$\rm \:  =  \:\dfrac{1}{b}\displaystyle\lim_{x \to 0} \frac{sinax}{x}$

$\rm \:  =  \:\dfrac{1}{b}\displaystyle\lim_{x \to 0} \frac{sinax}{ax} \times a$

$\rm \:  =  \:\dfrac{a}{b} \: \displaystyle\lim_{x \to 0} \frac{sinax}{ax}$

We know,

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

So, using this identity, we get

$\rm \:  =  \:\dfrac{a}{b} \times 1$

$\rm \:  =  \:\dfrac{a}{b}$

Hence,

$\green{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \bf \: \frac{sinax}{bx} = \frac{a}{b} \: }}}$

More to know -

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

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

$\green{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \bf \: \frac{1 - cosx}{x} = 0 \: }}}$

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

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

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