I have a mathematical problem with several parts, regarding the same function. Sometimes I have to prove continuity and sometimes to classify discontinuities.
I'll focus on the first section (Continuity on $x=0$) since it's virtually the same afterwards. The only limitation I have is that L'hopital's rule is not allowed to be used so no differentiation is allowed and everything is to be proved by the limits-methods. My calculations almost got me to the end but I feel like I'm missing 1 step before I solve it by myself, so I'd like your help.
Let there be a function $$f(x) = \left\{\begin{matrix} (1+ax+bx^2)^\frac{1}{sinx} & x>0\\ e^b & x=0\\ \frac{e^x-c}{sin(x)} & x<0 \end{matrix}\right.$$
We should find $a,b,c \in\mathbb{R}$ for which $f$ is continuous, meaning : $$\lim_{x\rightarrow 0^-}{f(x)} = \lim_{x\rightarrow 0^+}{f(x)} =\lim_{x\rightarrow 0}{f(x)} \in \mathbb{R}$$
- For starters I claimed $\lim_{x\rightarrow 0^+}{f(x)}\in \mathbb{R}$ - therefore:
$$\lim_{x\rightarrow 0}\Big({\frac{e^x-c}{sin(x)}\Big)} = \lim_{x\rightarrow 0}{\frac{e^x-c}{sin(x)}} \times\frac{x}{x} = \lim_{x\rightarrow 0}{\frac{e^x-c}{x} \times \Big(\frac{sin(x)}{x}}\Big)^{-1} = \lim_{x\rightarrow 0}{\frac{e^x-c}{x}}$$
Since $$\lim_{x\rightarrow 0}{\frac{e^x-c}{x}} \in\mathbb{R}$$ For any $c \neq 1$ the limit goes to $\infty$ Therefore $c=1$
- We claim continuity, therefore: $$\lim_{x\rightarrow 0^-}{f(x)} =\lim_{x\rightarrow 0}{f(x)}$$ $$\lim_{x\rightarrow 0}{(1+ax+bx^2)^{{sinx}^{-1}}} = e^b$$ I manipulated the power since : $\lim_{x\rightarrow 0}{\frac{sinx}{x}} = 1$ (this one is given) $$\frac{1}{sinx} =\frac{1}{sinx}\times\frac{x}{x} = \frac{1}{x}\times\Big(\frac{sinx}{x}\Big)^{-1} = \frac{1}{x}$$ Therefore: $$\lim_{x\rightarrow 0^-}{(1+ax+bx^2)^{{sinx}^{-1}}} = \lim_{x\rightarrow 0}{\Big(1+x(a+bx)\Big)^{\frac{1}{x}}} = \lim_{x\rightarrow 0}{e^{a+bx}} = e^a$$ I used Euler's number above: $\lim_{t\rightarrow 0}{(1 +t)^\frac{1}{t}} = e$
From continuity I calim that $$\lim_{x\rightarrow 0^-} = f(0) \rightarrow e^a = e^b \rightarrow a=b$$
- Now this is the part where I'm stuck. I somehow need to prove that $a=b=0$ but I couldn't find any. Since im 90% through I'd like to be pointed what I'm missing in order to solve this.
UPDATE:
since we couldn't have used differentiation this question was insolvable. we had to show that:
$$\lim_{x\rightarrow 0}{\frac{(e^x - 1)}{x}} = 1$$ since it's the very definition of the derivative of $e^x$: $$\lim_{x\rightarrow 0}{\frac{(e^x - 1)}{x}} = \lim_{x\rightarrow 0}{\frac{(e^x - e^0)}{x-0}} = \frac{d}{dx}{e^x} = e^x_{(x=0)} = 1$$
Thank you all.