Find the limit: mathop {lim }limits_{x to -1} | vedclass

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

Question

(B) The given expression is a polynomial function $f(x) = 1+x+x^{2}+ \dots +x^{10}$.Since polynomial functions are continuous everywhere,the limit as

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The given expression is a polynomial function $f(x) = 1+x+x^{2}+ \dots +x^{10}$.Since polynomial functions are continuous everywhere,the limit as $x 	o -1$ is simply the value of the function at $x = -1$.Substituting $x = -1$ into the expression:$f(-1) = 1 + (-1) + (-1)^{2} + (-1)^{3} + (-1)^{4} + (-1)^{5} + (-1)^{6} + (-1)^{7} + (-1)^{8} + (-1)^{9} + (-1)^{10}$$f(-1) = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1$Grouping the terms: $(1-1) + (1-1) + (1-1) + (1-1) + (1-1) + 1$$f(-1) = 0 + 0 + 0 + 0 + 0 + 1 = 1$Therefore,the limit is $1$.

Find the limit: $\mathop {\lim }\limits_{x \to -1} (1+x+x^{2}+ \dots +x^{10})$

Similar Questions

Let $x_n = (2^n + 3^n)^{\frac{1}{2n}}$ for all natural numbers $n$. Then,

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ is

If ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + \dots - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ then $\mathop {\lim }\limits_{n \to \infty } {x_n}$ is equal to

Let $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$. The smallest positive integer $k$ such that $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$ is

If the value of $\lim _{x \rightarrow 0}(2-\cos x \sqrt{\cos 2 x})^{\left(\frac{x+2}{x^{2}}\right)}$ is equal to $e^{a}$,then $a$ is equal to $.....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module