If mathop {lim }limits_{x to 1} frac{{{x^4} - | vedclass

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

Question

(B) First,evaluate the left-hand side limit:$\mathop {\lim }\limits_{x 	o 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x 	o 1} \frac{(x-

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(B) First,evaluate the left-hand side limit:$\mathop {\lim }\limits_{x 	o 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x 	o 1} \frac{(x-1)(x+1)(x^2+1)}{x-1} = \mathop {\lim }\limits_{x 	o 1} (x+1)(x^2+1) = (1+1)(1^2+1) = 2 	imes 2 = 4$.Next,evaluate the right-hand side limit:$\mathop {\lim }\limits_{x 	o k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \mathop {\lim }\limits_{x 	o k} \frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)} = \mathop {\lim }\limits_{x 	o k} \frac{x^2+xk+k^2}{x+k} = \frac{k^2+k^2+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2}$.Equating both sides:$4 = \frac{3k}{2}$ $\Rightarrow 3k = 8$ $\Rightarrow k = \frac{8}{3}$.

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$,then $k$ is

Similar Questions

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$,then $(a+b)$ is equal to

If $\lim _{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}$,then $2 \alpha-\beta$ is equal to :

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

If $\lim _{x \rightarrow 0} \frac{a x^2 e^x - b \log _e(1+x) + c x e^{-x}}{x^2 \sin x} = 1$,then $16(a^2 + b^2 + c^2)$ is equal to ...........................

If $n > 0$ and $\lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0$,then the minimum value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module