Q.Find $a$ and $b$ if the following function is continuous at the point indicated against it: $f(x) = \dfrac{4\tan x + 5\sin x}{a^x-1}$, for $x < 0$, $= \dfrac{9}{\log 2}$, for $x=0$, $= \dfrac{11x+7x\cos x}{b^x-1}$, for $x > 0$.

$f(x)=\dfrac{4\tan x+5\sin x}{a^x-1}$ for $x<0$, $f(0)=\dfrac{9}{\log2}$, $f(x)=\dfrac{11x+7x\cos x}{b^x-1}$ for $x>0$.

Left-hand limit: as $x\to0$, numerator $4\tan x+5\sin x\sim4x+5x=9x$ (using $\tan x\sim x,\sin x\sim x$), and denominator $a^x-1\sim x\log a$. So $\displaystyle\lim_{x\to0^-} f(x)=\dfrac{9x}{x\log a}=\dfrac{9}{\log a}$. Setting this equal to $f(0)=\dfrac{9}{\log2}$: $\dfrac{9}{\log a}=\dfrac{9}{\log2}\Rightarrow\log a=\log2\Rightarrow a=2$. …

Replace each numerator by its leading small-$x$ behaviour ($\tan x\sim x,\sin x\sim x,\cos x\to1$) and each denominator by $a^x-1\sim x\log a$, then solve …

Forgetting to factor out $x$ from $11x+7x\cos x=x(11+7\cos x)$ before taking the limit …