Question

lim (x--> 0) sin ax- sin bx/ sin cx - sin dx​

Answer 1

Evaluate

$\rm \:\lim_{x\to 0} \: \dfrac{sinax \: - \: sinbx}{sincx \: - \:sindx }$

$\large\underline\purple{\bold{Solution :- }}$

Concept used :-

Concept used :- L - Hospital Rule

• So, L'Hospital's Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

Now,

Consider,

$\rm \:\lim_{x\to 0} \: \dfrac{sinax \: - \: sinbx}{sincx \: - \:sindx }$

On substituting the value of x directly, we get indeterminant form

So, By L Hospital Rule, we have

$= \: \rm \:\lim_{x\to 0} \: \dfrac{\dfrac{d}{dx} (sinax \: - \: sinbx)}{\dfrac{d}{dx} (sincx \: - \:sindx) }$

$= \rm \:\lim_{x\to 0} \: \dfrac{a \: cosax \: - \: b \: cosbx}{c \: coscx \: - d\:cosdx }$

$\rm : \implies \: \dfrac{a \: - \: b}{c \: - \: d}$

$\large{\boxed{\boxed{\bf{Explore \ more : - }}}}$

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ log(1 + x) \: }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} \: - \: 1}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {a}^{x} \: - \: 1 }{x} \: = \: log(a) }}}}}} \\ \end{gathered}$