Question

Evaluate: ${lim}_{x\to 0}\frac{e^{\alpha x}-e^{\beta x}}{sin\alpha x-sin\beta x}$

Answer 1

${lim}_{x\to 0}\frac{e^{\alpha x}-e^{\beta x}}{sin\alpha x-sin\beta x}$

$={lim}_{x\to 0}\frac{e^{\alpha x}(1-\frac{e^{\beta x}}{\alpha x})}{2\cos \left(\frac{\alpha x+\beta x}{2}\right)\sin \left(\frac{\alpha x-\beta x}{2}\right)}$ .........$\left(1\right)$ using transformation angle formula $\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

Using the formulae, ${lim}_{x\to 0}\frac{e^x-1}{x}=1$

${lim}_{x\to 0}\cos \theta =1$ and ${lim}_{\theta \to 0}\frac{\sin \theta }{\theta }=1$

we evaluate the above expression as

${lim}_{x\to 0}{e}^{\alpha x}=1$

${lim}_{x\to 0}(1-e^{(\beta -\alpha )x})={lim}_{x\to 0}\frac{(1-e^{(\beta -\alpha )x})}{(\beta -\alpha )x}\times (\beta -\alpha )x=-1\times {lim}_{x\to 0}(\beta -\alpha )x$

${lim}_{x\to 0}\cos \left(\frac{\alpha +\beta }{2}\right)x=1$

${lim}_{x\to 0}\sin \left(\frac{\alpha -\beta }{2}\right)x={lim}_{x\to 0}\frac{\sin \left(\frac{\alpha -\beta }{2}\right)x}{\left(\frac{\alpha -\beta }{2}\right)x}\times \left(\frac{\alpha -\beta }{2}\right)x={lim}_{x\to 0}\left(\frac{\alpha -\beta }{2}\right)x$

Substituting the above in $\left(1\right)$ we get

$=\frac{-1\times {lim}_{x\to 0}(\beta -\alpha )x}{2\times 1\times {lim}_{x\to 0}\left(\frac{\alpha -\beta }{2}\right)x}$ the term $x$ gets cancelled in both the numerator and the denominator

$=\frac{-\beta +\alpha }{\alpha -\beta }=1$