Question

evaluate limit X tends to 0 (cosax-cosbx)/x^2​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \dfrac{cosax - cosbx}{ {x}^{2} }$

If we put directly x = 0, we get

$\rm \:  =  \:  \: \dfrac{cos0 - cos0}{ {0}^{2} }$

$\rm \:  =  \:  \: \dfrac{0}{0}$

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

So, to evaluate the limit of the above function,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \dfrac{cosax - cosbx}{ {x}^{2} }$

We know,

$\boxed{ \bf{ \: cosx - cosy = 2sin\bigg(\dfrac{x + y}{2} \bigg) \: sin\bigg(\dfrac{y - x}{2} \bigg)}}$

So, using this result, we get

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0} \dfrac{2sin\bigg(\dfrac{ax + bx}{2} \bigg) \: sin\bigg(\dfrac{bx - ax}{2} \bigg)}{ {x}^{2} }$

can be rewritten as

$\rm \:  =  \:  \: 2\displaystyle\lim_{x \to 0} \dfrac{sin\bigg(\dfrac{a+ b}{2} \bigg)x}{x} \times \displaystyle\lim_{x \to 0} \dfrac{sin\bigg(\dfrac{b - a}{2} \bigg)x}{x}$

$\rm = 2\displaystyle\lim_{x \to 0} \dfrac{sin\bigg(\dfrac{a+ b}{2} \bigg)x}{\bigg(\dfrac{a + b}{2} \bigg)x} \times \bigg(\dfrac{a + b}{2} \bigg) \times \displaystyle\lim_{x \to 0} \dfrac{sin\bigg(\dfrac{b - a}{2} \bigg)x}{\bigg(\dfrac{b - a}{2} \bigg)x} \times \bigg(\dfrac{b - a}{2} \bigg)$

We know,

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

So, using this result, we get

$\rm \:  =  \:  \: 2 \times \bigg(\dfrac{a + b}{2} \bigg) \times \bigg(\dfrac{b - a}{2} \bigg)$

$\rm \:  =  \:  \: \dfrac{ {b}^{2} - {a}^{2} }{2}$

Hence,

$\underbrace{ \boxed{\:\displaystyle\lim_{x \to 0} \bf \: \: \: \dfrac{cosax \: - \: cosbx}{ {x}^{2} } \: = \: \frac{ {b}^{2} \: - \: {a}^{2} }{2} \: \: \: }}$

Additional Information :-

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

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

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

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

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

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