Limit Continuity and Differentiability 4 Question 4
4. $\operatorname{Let} f(x)=\lim _{n \rightarrow \infty} \frac{n^{n}(x+n) x+\frac{n}{2} \ldots x+\frac{n}{n}}{n !\left(x^{2}+n^{2}\right) x^{2}+\frac{n^{2}}{4} \ldots x^{2}+\frac{n^{2}}{n^{2}}}$,
for all $x=0$. Then
(a) $f \frac{1}{2} \geq f(1)$
(b) $f \frac{1}{3} \leq f \frac{2}{3}$
(c) $f^{\prime}(2) \leq 0$
(d) $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$
(2016 Adv.)
Numerical Value
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Answer:
Correct Answer: 4. (c)
Solution:
- Given function $f(x)=[x]-\frac{x}{4}, x \in R$
Now, $\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0}[4+h]-\frac{4+h}{4}$
$\left[\because\right.$ put $x=4+h$, when $x \rightarrow 4^{+}$, then $h \rightarrow 0$ ]
$$ =\lim _{h \rightarrow 0}(4-1)=3 $$
and $\lim _{x \rightarrow 4^{-}} f(x)=\lim _{h \rightarrow 0}[4-h]-\frac{4-h}{4}$
$\left[\because\right.$ put $x=4-h$, when $x \rightarrow 4^{-}$
then $h \rightarrow 0]$
$$ =\lim _{h \rightarrow 0}(3-0)=3 $$
and $f(4)=[4]-\frac{4}{4}=4-1=3$
$\because \lim _{x \rightarrow 4^{-}} f(x)=f(4)=\lim _{x \rightarrow 4^{+}} f(x)=3$
So, function $f(x)$ is continuous at $x=4$.
