If $\begin{aligned} f(x) &=\frac{\left(e^{3 x}-1\right) \sin x^{\circ}}{x^{2}} \text { if } x \neq 0 \\…
If $\begin{aligned} f(x) &=\frac{\left(e^{3 x}-1\right) \sin x^{\circ}}{x^{2}} \text { if } x \neq 0 \\ &=\frac{\pi}{60} \quad \text { if } x=0, \text { then } \end{aligned}$
- $f$ is continuous at $x=0$
- $\lim _{x \rightarrow 0} f(x)=3$
- $f$ has irremovable discontinuity at $x=0$
- $f$ has removable discontinuity at $x=0$
Solution
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right) \sin x^{0}}{x^{2}}$
Dividing numerator and denominator by $x^{2}$, we get
$\begin{array}{l}
=\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{3 x} \times 3\right) \frac{\sin \left(\frac{\pi x}{180}\right)^{0}}{\left(\frac{\pi x}{180}\right)^{c}} \times\left(\frac{\pi}{180}\right) \\
=\frac{\left(\frac{x^{2}}{x^{2}}\right)}{=(3)(1)\left(\frac{\pi}{180}\right)}=\frac{\pi}{60}
\end{array}$
Hence $f(x)$ is continuous at $x=0$.
Asked in: MHT CET 2020 (20 Oct Shift 2)
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