Question

$f(x)=\begin{cases} 3-\left[\cot^{-1}\dfrac{2x^3-3}{x^2}\right] & x>0\\\\ \{x^2\} \cos e^{\frac{1}{x}} & x<0\end{cases}$

What is the value of f(0), if f(x) is continuous at x=0?
( [...] is G.I.F. and {...} is Functional Part Function)

NO SPAMMING. ONLY CORRECT ANSWERS AND WITH FULL EXPLANATION.

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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$f(x)=\begin{cases} 3-\left[\cot^{-1}\dfrac{2x^3-3}{x^2}\right] & x>0\\\\ \{x^2\} \cos e^{\frac{1}{x}} & x<0\end{cases}$

It is further given that f(x) is continuous at x = 0.

We know,

A function f(x) is said to continuous at x = a iff

$\boxed{ \rm{ \displaystyle\lim_{x \to a^-} \: f(x) = \displaystyle\lim_{x \to a^+} \: f(x) = \: f(a)}}$

So, here it is given that, f(x) is continuous at x = 0

$\bf\implies \:{ \rm{ \displaystyle\lim_{x \to 0^-} \: f(x) = \displaystyle\lim_{x \to 0^+} \: f(x) = \: f(0)}}$

So,

Let evaluate LHL

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0^-} \: f(x)}$

$\rm \:  =  \:  \: \displaystyle\lim_{x \to 0^-} \: \{x^2\} \cos e^{\frac{1}{x}}$

$\red{\rm :\longmapsto\:Put \: x = 0 - h, \: as \: x \to 0 \: so \: h \to 0}$

So, above expression can be rewritten as

$\rm \:  =  \:  \: \displaystyle\lim_{h \to 0}\{(0 - h)^2\} \cos e^{\frac{1}{(0 - h)}}$

$\rm \:  =  \:  \: \displaystyle\lim_{h \to 0}\{(- h)^2\} \cos e^{\frac{1}{(- h)}}$

$\rm \:  =  \:  \: \displaystyle\lim_{h \to 0}\{h^2\} \cos e^{\frac{1}{(- h)}}$

$\rm \:  =  \:  \: 0 \times cos {e}^{ - \infty }$

$\rm \:  =  \:  \: 0 \times cos0$

$\rm \:  =  \:  \: 0 \times 1$

$\rm \:  =  \:  \: 0$

$\bf\implies \:\displaystyle\lim_{x \to 0^-} \: \{x^2\} \cos e^{\frac{1}{x}} \: = \: 0$

Since,

$\red{\rm :\longmapsto\:f(0) = \displaystyle\lim_{x \to 0^-} \: f(x)}$

$\red{\rm :\longmapsto\:f(0) = \displaystyle\lim_{x \to 0^-} \: \{x^2\} \cos e^{\frac{1}{x}}}$

$\red{\rm :\longmapsto\:f(0) = 0}$

Additional Information :-

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \: \frac{sinx}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \: \frac{tanx}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \: \frac{tan {}^{ - 1} x}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \: \frac{sin {}^{ - 1} x}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \: \frac{log(1 + x)}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \: \frac{ {e}^{x} - 1}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \: \frac{ {a}^{x} - 1}{x} = loga}}$