Determine the values of $a,b,c$ for which the function is continuous at $x=0$

Question

Determine the values of $a,b,c$ for which the function is continuous at $x=0$
$$ f(x) = \begin{cases} \frac{\sin(a+1)x+sinx}{x} \qquad \text{if} \ x<0 \ ; \\ \\ c \qquad\quad\qquad \text{if} \ x=0 \ ; \\ \\ \frac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}} \ \ \ \text{if} \ x>0 \ . \end{cases} $$

I tried to solve the problem like this: $$\lim_{x\to 0}\frac{\sin(a+1)x+sinx}{x}$$
$$=\lim_{x\to 0}\frac{acosx}{x}+cosax +1$$
$$=2+a$$ and $$\lim_{x\to 0}\frac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}}$$
$$\lim_{x\to 0}\frac{\sqrt{1+bx}-1}{bx}$$
$$=?$$ Can you help me solve this problem?

For the last limit, this is just the definition of a derivative at $x=0$ where $g(x)=\sqrt{1+bx}$. — Clayton, Apr 19 '13 at 11:46
Take the derivative and evaluate it at $x=0$, then multiply by $1/b$. What do you get? — Clayton, Apr 19 '13 at 11:52

score 2 · Answer 1 · answered Apr 19 '13 at 11:54

2

Hint: $$\frac{\sqrt{1+bx}-1}{bx}=\frac{\sqrt{1+bx}-1}{bx}\cdot\frac{\sqrt{1+bx}+1}{\sqrt{1+bx}+1}=\frac{1}{\sqrt{1+bx}+1}$$

answered Apr 19 '13 at 11:54

Easy

4,485

Okay the limit is 1/2 and how can i find b? – chndn Apr 19 '13 at 11:57
so how can i find b? 2+a=c=1/2 so where is b? – chndn Apr 19 '13 at 11:59
@chndn, I don't see any condition for $b$ except $x+bx^2\geq0$. – Easy Apr 19 '13 at 12:08

Determine the values of $a,b,c$ for which the function is continuous at $x=0$

1 Answers1