If f(x) = begin{cases} frac{2}{5-x}, & x

Question

(C) To find the limits at $x = 3$,we evaluate the left-hand limit and right-hand limit separately.For the right-hand limit $(x 	o 3^+)$,we use the de

Vedclass · Accepted Answer

$\lim_{x 	o 3^+} f(x) 
eq \lim_{x 	o 3^-} f(x)$

Vedclass · Answer

(C) To find the limits at $x = 3$,we evaluate the left-hand limit and right-hand limit separately.For the right-hand limit $(x 	o 3^+)$,we use the definition $f(x) = 5 - x$:$\lim_{x 	o 3^+} f(x) = 5 - 3 = 2$.For the left-hand limit $(x 	o 3^-)$,we use the definition $f(x) = \frac{2}{5-x}$:$\lim_{x 	o 3^-} f(x) = \frac{2}{5 - 3} = \frac{2}{2} = 1$.Since $2 
eq 1$,we have $\lim_{x 	o 3^+} f(x) 
eq \lim_{x 	o 3^-} f(x)$.

If $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x > 3 \end{cases}$,then:

Similar Questions

If $\alpha, \beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$,then $\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$ equals

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{ax + x \cos x}{b \sin x}$

$\lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right] = $

The value of $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$ is

$\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module