Question

f(x)= [sin(a+1)x + sinx /x , x c , x=0

√x +bx^2 - √x / b x^3/2 , x >0


function is continuous at x=0

Answer 1

$\bold{F(x) = \frac{sin(a+1)x + sinx}{x}}$ at x < 0
take limit ,
F(0⁻) = $\lim_{x \to 0^-} \bold{ \frac{sin(a+1)x + sinx}{x}}$
we see the form of limit is 0/0, so, Use L-Hospital rule for it ,
F(0⁻) = $\lim_{x \to 0^-} \bold{ \frac{cos(a+1)x.(a+1) + cosx}{1}}$
F(0⁻) = (a + 1) + 1 = (a + 2) -----------(1)

F(x) = c at x = 0 , then F(0) = c ---------(2)

F(x) = $\bold{\frac{\sqrt{x +bx^2}-\sqrt{x}}{bx^{3/2}}}$ at x > 0
Then take limit ,
F(0⁺) = $\lim_{x \to 0^+} \bold{\frac{\sqrt{x +bx^2}-\sqrt{x}}{bx^{3/2}}}$
= $\lim_{x \to 0^+} \bold{\frac{\sqrt{1+bx}-1}{bx}}$
This is like the standard form of limit solution ,
So, F(0⁺) = 1/2

F(x) will be continuous at x = 0 only when F(0⁻) = F(0⁺) = F(0)
a + 2 = c = 1/2
⇒c = 1/2
⇒ a = -3/2
⇒ b∈ R

Hence, a = -3/2 , c = 1/2 and b∈R

Answer 2

Step-by-step explanation:

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