Limit with 3-unknown variable.

[man111]

find real no. [tex]\alpha,\beta[/tex] and [tex]\gamma[/tex] such that [tex]\lim_{x \to 0}\frac{e^{6x}+\alpha x+\beta}{x^\gamma} = 1[/tex]

[mkmarinov]

1) Let [tex]L=\lim_{x \to 0} \frac{e^{6x}+\alpha x + \beta}{x^\gamma}[/tex] be an indeterminate form of the kind [tex]\frac{0}{0}[/tex]; i.e. [tex]\gamma > 0[/tex] and the numerator is 0.
Using L'hospital's rule:
[tex]\lim_{x \to 0} \frac{e^{6x}+\alpha x + \beta}{x^\gamma}=\lim_{x \to 0} \frac{6e^{6x}+\alpha}{\gamma x^{\gamma -1}}[/tex]
1) a) let [tex]\gamma > 1[/tex]. Using L'hospital's again: [tex]L=\lim_{x \to 0}\frac{36e^{6x}}{\gamma(\gamma-1)x^{\gamma-2}}=[\frac{1}{0}][/tex]
1) b) now let [tex]\gamma = 1[/tex]. [tex]L=\lim_{x \to 0}\frac{e^{6x}+\alpha x + \beta}{x}=\lim_{x \to 0}(6e^{6x}+\alpha)=6+\alpha=1 => \alpha=-5[/tex].
But the numerator is zero => [tex]e^{6.0}-5.0+\beta=0 => \beta=-1[/tex].
So, one solution is [tex]\alpha = -5; \beta = -1; \gamma = 1[/tex].
1) c) now [tex]\gamma \in (0;1)[/tex]. [tex]L=\lim_{x \to 0} \frac{x^{1-\gamma}(6e^{6x}+\alpha)}{\gamma}=0(\frac{6+\alpha}{\gamma})=0[/tex], so there are no solutions for this case.
2) Let [tex]\gamma = 0[/tex]
[tex]L=\lim_{x \to 0} (e^{6x}+\alpha x + \beta)=1+0+\beta = 1 => \beta = 0[/tex]. Another solution is:
[tex]any \alpha ; \beta = 0; \gamma = 0[/tex]
3) Let [tex]\gamma<0[/tex] - the limit is obviously zero.

So, solutions are:
[tex](-5; -1; 1)\cup (\alpha ; 0 ; 0)[/tex]

[man111]

Thanks mkmarinov for nice case study. as well as explanation.