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Given that $f(x)= \dfrac{\sin (a+1)x+ \sin x} {x}$ if $x<0.$

$$f(x)= c \text{ if } x=0.$$

$$f(x)= \frac{{\sqrt{x+bx^2}}-\sqrt {x}}{b \sqrt{x^3}} \text{ if } x>0.$$

Also given that the function is continuous at $x=0.$ Then find $a,b,c.$ I tried simplifying the left hand limit of the function and equated it to $f(0)$ and ended up with $a-c=-2$ Now i am having problem simplifying the right hand limit of the function.Please help. Also after finding the left hand limit and equation it to $f(0)$ how should i proceed to find the values required?

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    $\begingroup$ Is it $\sin((a+1)x)$ or $\sin(a+1) \cdot x$? $\endgroup$ – the_candyman Nov 6 '17 at 16:05
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To find limit of: $$f(x)=\frac{\sqrt{x+bx^2}-\sqrt{x}}{b\sqrt{x^3}}=\frac{\sqrt{1+bx}-1}{bx}$$ we set $p(x)=\sqrt{1+bx}-1$ and $q(x)=bx$. Thus $f(x)=\frac{p(x)}{q(x)}$

Since $lim_{x\to 0} p(x)=lim_{x\to 0} q(x)=0 $ we can apply to L'Hospital's rule to get: $$lim_{x\to 0} f(x)=lim_{x\to 0} \frac{p'(x)}{q'(x)}=lim_{x\to 0} \frac{\frac{b}{2}(1+bx)^{-\frac{1}{2}}}{b}=\frac{1}{2}$$

From this we can conclude that $c=\frac{1}{2}$

Now we can express $a$ as $a=c-2=-\frac{3}{2}$ while $b$ may take any real value.

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