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

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $

$\mathop {Lim}\limits_{x \to c} f(x)$ does not exist when: (where $[x]$ denotes the greatest integer function and $\{x\}$ denotes the fractional part function.)

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}} = $

$\lim \limits_{x \rightarrow 0} \left(\tan \left(\frac{\pi}{4}+x\right)\right)^{\frac{1}{x}}$ is equal to

If $f(x) = \begin{cases} \frac{\sin([x])}{[x]}, & \text{when } [x] \neq 0 \\ 0, & \text{when } [x] = 0 \end{cases}$ where $[x]$ is the greatest integer function,then $\lim_{x \to 0} f(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module