NIMCET 2024 — Mathematics PYQ

Continuity at  x=0:

A function $f(x)$ is continuous at $x=0$ if
$$<br>\underset{x\to 0}{\lim }f(x)=f(0).<br>$$

We are given
<br>f(0)=\frac{1}{\pi}, \qquad&nbsp;<br>f(x)=\frac{x}{e^{\pi x}-1} \quad (x\neq 0).<br>

Evaluate the limit:
$$<br>\underset{x\to 0}{\lim }\frac{x}{e^{\pi x}-1}<br>=\frac{1}{\pi }\underset{x\to 0}{\lim }\frac{\pi x}{e^{\pi x}-1}.<br>$$

Let $u=\pi x$. Then as $x\to 0$, $u\to 0$, and we use the standard limit
$$<br>\underset{u\to 0}{\lim }\frac{e^u-1}{u}=1.<br>$$

Hence,
$$<br>\underset{x\to 0}{\lim }f(x)<br>=\frac{1}{\pi }\underset{u\to 0}{\lim }\frac{u}{e^u-1}<br>=\frac{1}{\pi }.<br>$$

Since
$$<br>\underset{x\to 0}{\lim }f(x)=f(0),<br>$$
the function is continuous at $x=0$.

Differentiability at  x=0:

We use the definition of derivative:
$$<br>f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{f(h)-f(0)}{h}.<br>$$

Substitute the given function:
<br>f'(0)<br>=\lim_{h\to 0}&nbsp;<br>\frac{\frac{h}{e^{\pi h}-1}-\frac{1}{\pi}}{h}.<br>

Simplify:
$$<br>f^{\prime }(0)<br>=\underset{h\to 0}{\lim }<br>\left(<br>\frac{1}{e^{\pi h}-1}-\frac{1}{\pi h}<br>\right).<br>$$

Use the expansion
$$<br>e^{\pi h}-1=\pi h+\frac{(\pi h{)}^2}{2}+\cdots ,<br>$$
which gives
$$<br>\frac{1}{e^{\pi h}-1}<br>=\frac{1}{\pi h}\left(1-\frac{\pi h}{2}+\cdots \right).<br>$$

Thus,
$$<br>f^{\prime }(0)<br>=-\frac{1}{2}.<br>$$

Final results:
$$<br>\text{The function is continuous at }x=0.<br>$$
$$<br>f^{\prime }(0)=-\frac{1}{2}.<br>$$