Question

If $\left \begin{array}{ll} f(x)= \frac{sin 4x}{5x}+a , & \quad for \ x \ \textgreater \ 0 \\ \hspace{0.75cm}= x+4-b , & \quad for \ x \ \textless \ 0 \\ \hspace{0.75cm}= 1 , & \quad for \ x = 0 \end{array} \right$ is continuous at x=0, find a and b.

Answer 1

any function, y = f(x) will be continuous at x = a only when,

$\displaystyle\lim_{x\to a^+}f(x)=\displaystyle\lim_{x\to a^-}f(x)=f(a)$

it is given that, function is continuous at x = 0,

so, $\displaystyle\lim_{x\to 0^+}f(x)=\displaystyle\lim_{x\to 0^-}f(x)=f(0)$

here, for x > 0 , f(x) = sin4x/5x + a

applying $\displaystyle\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{sin4x}{5x}+a$

= 4/5 + a

for x < 0, f(x) = x + 4 - b

so, $\displaystyle\lim_{x\to 0^-}(x+4-b)$

= 4 - b

and at x = 0 , f(x) = 1

so, 4/5 + a = 4 - b = 1

so, 4/5 + a = 1 ⇒a = 1/5

and 4 - b = 1 ⇒b = 3

hence, a = 1/5 and b = 3

Answer 2

your solution is attached above ...