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:

  1. 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 $x=4$.



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