GATE 2024 DA Q.60: solution and answer

Official answer: 0.5Source: Official answer key, GATE 2024 DA (IISc Bengaluru)

1. Near $x=0$ both $x^2+1$ and $\cos x$ are positive, so the logarithm of the product splits: $\ln ((x^2+1)\cos x)=\ln (1+x^2)+\ln (\cos x)$.

2. Divide by $x^2$: the limit is ${\lim }_{x\to 0}\frac{\ln (1+x^2)}{x^2}+{\lim }_{x\to 0}\frac{\ln (\cos x)}{x^2}$, valid provided both limits exist (shown below).

3. First term: put $t=x^2\to 0^{+}$. Then $\frac{\ln (1+t)}{t}\to 1$, since $\ln (1+t)=t-\frac{t^2}{2}+\cdots$ (equivalently, the derivative of $\ln (1+t)$ at $t=0$ is $1$).

4. Second term: $\cos x=1-\frac{x^2}{2}+O(x^4)$. With $u=-\frac{x^2}{2}+O(x^4)\to 0$, $\ln (1+u)=u+O(u^2)$, so $\ln \cos x=-\frac{x^2}{2}+O(x^4)$. Therefore $\frac{\ln \cos x}{x^2}\to -\frac{1}{2}$.

5. Check by L'Hopital (the form is $0/0$): $\frac{d}{dx}\ln \cos x=-\tan x$ and $\frac{d}{dx}{x}^2=2x$, so the ratio is $-\frac{\tan x}{2x}\to -\frac{1}{2}$.

6. Both limits exist, so the sum is $1-\frac{1}{2}=\frac{1}{2}=0.5$.