[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \lim_{x \to0} \frac{e^{ax} -e^{bx} }{ \sin\:ax- \sin\:bx} \\ [/tex]

Pra resolver esse limite, basta aplica a regra de L'Hopital, uma vez que ao substituir o valor a qual o "x" tende, obtém-se 0/0.

• A regra de L'Hopital nos diz que:

[tex]\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \: \: ou \: \: \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{ \infty }{ \infty } \\ \\ \lim_{x \to a} \frac{ \frac{d}{dx} f(x)}{ \frac{d}{dx} g(x)} [/tex]

Ou seja, em caso como esses citados acima, basta derivar o numerador e o denominador.

[tex] \lim_{x \to0} \frac{e^{ax} -e^{bx} }{ \sin ax- \sin bx} = \lim_{x \to0} \frac{ \frac{d}{dx} (e^{ax} -e^{bx}) }{ \frac{d}{dx} (\sin ax- \sin bx)} \\ \\ \lim_{x \to0}\frac{ae^{ax} -be^{bx} }{a \cos\:ax- b\cos\:bx}[/tex]

Tendo feito a derivação, agora é só substituir o valor a qual o "x" tende:

[tex]\lim_{x \to0} \frac{a.e {}^{a.0} - b.e {}^{b.0} }{a \cos(a.0) - b. \cos(b.0)} = \lim_{x \to0} \frac{a.e {}^{0} - b.e {}^{0} }{a \cos(0) - b. \cos(0)} \\ \\ \lim_{x \to0} \frac{a.1 - b.1 }{a.1 - b. 1} =\lim_{x \to0} 1 = \boxed{ 1 } \\ [/tex]

[tex]\displaystyle \sf \lim_{x\to 0 } \left(\frac{e^{ax}-e^{bx}}{sen(ax)-sen(bx)}\right) \\\\\\ \text{Resolvendo sem usar a regra de L'H\^opital...}\\\\ \text{s\'o me vem na mente as s\'eries de Taylor e Mclaurin..} :\\\\ e^{x} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+ \frac{x^4}{4!}+... \\\\ \text{Fa\c camos} :\\\\ e^{ax} = 1+(ax)+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+ \frac{(ax)^4}{4!}+...[/tex]

[tex]\displaystyle \sf e^{bx} = 1+(bx)+\frac{(bx)^2}{2!}+\frac{(bx)^3}{3!}+ \frac{(bx)^4}{4!} +...\\\\\\ e^{ax}-e^{bx} = (ax)-(bx)+\frac{(ax)^2-(bx)^2}{2!}+\frac{(ax)^3-(bx)^3}{3!}+ \frac{(ax)^4-(bx)^4}{4!}+...[/tex]

Do denominador :

[tex]\displaystyle \sf sen(x) =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+... \\\\\\ sen(ax) = (ax)-\frac{(ax)^3}{3!}+\frac{(ax)^5}{5!}-\frac{(ax)^7}{7!}+... \\\\\\ sen(bx) = (bx)-\frac{(bx)^3}{3!}+\frac{(bx)^5}{5!}-\frac{(bx)^7}{7!}+... \\\\\\ da\'i }:\\\\ sen(ax)-sen(bx) = (ax)-(bx)-\frac{\left[(ax)^3-(bx)^3\right]}{3!}+\frac{(ax)^5-(bx)^5 }{5!}-...[/tex]

Assim,

[tex]\displaystyle \sf \lim_{x\to 0 } \left(\frac{e^{ax}-e^{bx}}{sen(ax)-sen(bx)}\right) \\\\\\\ \sf \lim_{x\to 0 } \left(\frac{\displaystyle (ax)-(bx)+\frac{(ax)^2-(bx)^2}{2!}+\frac{(ax)^3-(bx)^3}{3!}+...}{\displaystyle (ax)-(bx)-\frac{\left[(ax)^3-(bx)^3\right]}{3!}+\frac{(ax)^5-(bx)^5 }{5!}-... }\right)[/tex]

Observe que tanto no numerador quanto no denominador o termo que podemos evidenciar é [ax-bx]  

[tex]\sf (ax)^2-(bx)^2= (ax-bx)(ax+bx) \\\\ (ax)^3-(bx)^3=(ax-bx)( (ax)^2+(ax)(bx)+(bx)^2) \\\\\ (ax)^4-(bx)^4= (ax-bx)^2(ax+bx)^2 \\\\ (ax)^5-(bx)^5= (ax-bx)((ax)^4+(ax)^3(bx)+...+(bx)^4)\\\ . ..[/tex]

evidenciando (ax-bx) :

[tex]\displaystyle \sf \lim_{x\to 0 } \left(\frac{\displaystyle [ax-bx]\left[1+\frac{[ax+bx]}{2!}+\frac{[(ax)^2+(ax)(bx)+(bx)^2]}{3!}+...\right]}{\displaystyle [ax-bx]\left[1-\frac{\left[(ax)^2+(ax)(bx)+(bx)^2\right]}{3!}+\frac{[(ax)^4+..]}{5!}-... \right]}\right)[/tex]

[tex]\displaystyle \sf \lim_{x\to 0 } \left(\frac{\displaystyle 1+\frac{[ax+bx]}{2!}+\frac{[(ax)^2+(ax)(bx)+(bx)^2]}{3!}+...}{\displaystyle 1-\frac{\left[(ax)^2+(ax)(bx)+(bx)^2\right]}{3!}+\frac{[(ax)^4+..]}{5!}-... }\right)[/tex]

[tex]\displaystyle \sf \left(\frac{\displaystyle 1+\frac{0}{2!}+\frac{0}{3!}+...}{\displaystyle 1-\frac{0}{3!}+\frac{0}{5!}-... }\right) = \frac{1}{1} = 1[/tex]

Portanto :

[tex]\displaystyle \sf \large\boxed{\sf \ \lim_{x\to 0 } \left(\frac{e^{ax}-e^{bx}}{sen(ax)-sen(bx)}\right) = 1 \ }\checkmark[/tex]