If f: R to [0, infty) is such that lim_{x to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $\lim_{x 	o 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$. Let $L = \lim_{x 	o 5} f(x)$. Since the limit exists,we can write $\lim_{x \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given that $\lim_{x 	o 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$. Let $L = \lim_{x 	o 5} f(x)$. Since the limit exists,we can write $\lim_{x 	o 5} ((f(x))^2 - 9) = L^2 - 9$. If $L^2 - 9 
eq 0$,then the limit $\lim_{x 	o 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}}$ would be of the form $\frac{k}{0}$ (where $k 
eq 0$),which would be $\infty$. Since the given limit is $0$,the numerator must approach $0$ as $x 	o 5$. Therefore,$L^2 - 9 = 0$,which implies $L^2 = 9$. Since the codomain of $f$ is $[0, \infty)$,$L$ must be non-negative. Thus,$L = 3$.

If $f: R \to [0, \infty)$ is such that $\lim_{x \to 5} f(x)$ exists and $\lim_{x \to 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$,then $\lim_{x \to 5} f(x)$ equals:

Similar Questions

$\lim _{x}$ ${\rightarrow -a} \frac{x^7+a^7}{x+a} = 7$ $\Rightarrow a = ?$

If $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$,then $P=$

If $f(x) = \begin{cases} 4x-5, & x \leq 2 \\ x-k, & x > 2 \end{cases}$ then the value of $k$ for which $\lim_{x \rightarrow 2} f(x)$ exists is equal to:

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0$, where $a, b \in R$, then:

If $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$ exists and is equal to $1$, then the value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module