ધારો કે $a_{1}, a_{2}, \dots, a_{n}$ એ નિશ્ચિત વાસ્તવિક સંખ્યાઓ છે અને વિધેય $f(x) = (x - a_{1})(x - a_{2}) \dots (x - a_{n})$ વ્યાખ્યાયિત કરો. $\lim_{x \to a_{1}} f(x)$ શું છે? કોઈ $a \neq a_{1}, a_{2}, \dots, a_{n}$ માટે,$\lim_{x \to a} f(x)$ ની ગણતરી કરો.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
આપેલ વિધેય $f(x) = (x - a_{1})(x - a_{2}) \dots (x - a_{n})$ છે.
$\lim_{x \to a_{1}} f(x) = \lim_{x \to a_{1}} [(x - a_{1})(x - a_{2}) \dots (x - a_{n})]$
$= (a_{1} - a_{1})(a_{1} - a_{2}) \dots (a_{1} - a_{n}) = 0 \times (a_{1} - a_{2}) \dots (a_{1} - a_{n}) = 0$.
તેથી,$\lim_{x \to a_{1}} f(x) = 0$.
હવે,$a \neq a_{1}, a_{2}, \dots, a_{n}$ માટે,
$\lim_{x \to a} f(x) = \lim_{x \to a} [(x - a_{1})(x - a_{2}) \dots (x - a_{n})]$
$= (a - a_{1})(a - a_{2}) \dots (a - a_{n})$.
તેથી,$\lim_{x \to a} f(x) = (a - a_{1})(a - a_{2}) \dots (a - a_{n})$.

Explore More

Similar Questions

$\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)(n^2+3 n-1)}$

$\lim _{x \rightarrow 0} x^3 \left\{ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right\} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{y^2}}}{x}$ ની કિંમત શોધો,જ્યાં ${y^2} = ax + b{x^2} + c{x^3}$.

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo