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If $f(x)=\left\{\begin{array}{cc}(1+|\sin x|)^{\frac{a}{|\sin x|}}, & -\pi / 6 < x < 0 \\ b & x=0 \\ \frac{\tan 2 x}{e^{\tan 3 x}}, & 0 < x < \pi / 6\end{array}\right.$
is continuous at $x=0$, find the values of $a$ and $b$.
Options:
is continuous at $x=0$, find the values of $a$ and $b$.
Solution:
1899 Upvotes
Verified Answer
The correct answer is:
$2 / 3, e^{2 / 3}$
We have, $\lim _{x \rightarrow 0^{-}} f(x)$
$$
\begin{aligned}
=& \lim _{x \rightarrow 0}\{1+|\sin x|\}^{\frac{a}{|\sin x|}} \\
=& \lim _{x \rightarrow 0}|\sin x| \cdot \frac{a}{|\sin x|}=e^{a} \\
\text { and } \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{\frac{\tan 2 x}{\tan 3 x}} \\
=& \lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x} \cdot \frac{3 x}{\tan 3 x} \times \frac{2}{3} \\
=& e^{2 / 3}
\end{aligned}
$$
For $f(x)$ to be continuous at $x=0$, we must have
$$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{x \rightarrow 0^{+}} f(x) \\
& &=f(0) \\
\Rightarrow & e^{a} &=e^{2 / 3}=b \\
\Rightarrow & a &=2 / 3 \\
\text { and } & b &=e^{2 / 3}
\end{aligned}
$$
$$
\begin{aligned}
=& \lim _{x \rightarrow 0}\{1+|\sin x|\}^{\frac{a}{|\sin x|}} \\
=& \lim _{x \rightarrow 0}|\sin x| \cdot \frac{a}{|\sin x|}=e^{a} \\
\text { and } \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{\frac{\tan 2 x}{\tan 3 x}} \\
=& \lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x} \cdot \frac{3 x}{\tan 3 x} \times \frac{2}{3} \\
=& e^{2 / 3}
\end{aligned}
$$
For $f(x)$ to be continuous at $x=0$, we must have
$$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{x \rightarrow 0^{+}} f(x) \\
& &=f(0) \\
\Rightarrow & e^{a} &=e^{2 / 3}=b \\
\Rightarrow & a &=2 / 3 \\
\text { and } & b &=e^{2 / 3}
\end{aligned}
$$
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