JEECUP (UP Polytechnic) Limits and Continuity — practice questions
32 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice JEECUP (UP Polytechnic) Limits and Continuity in the app →Evaluate $\displaystyle\lim_{x \to 0} \dfrac{ in 3x}{x}$.Evaluate $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$.Evaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.Evaluate $\displaystyle\lim_{x \to \infty} \left(1 + \dfrac{2}{x}\right)^x$.Define $f(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 2$. Is $f$ continuous at $x = 1$?Consider $f(x) = |x|$ at $x = 0$. Which statement is correct?lim_(x→2) (x² - 4)/(x - 2) =lim_(x→0) sin(x)/x =For limit to EXIST at x = a, left-hand and right-hand limits must:A function f is continuous at x = a if:d/dx (xⁿ) = ?lim_(x→0) (1 - cos x)/x² =lim_(x→∞) (1 + 1/x)^x =lim_(x→0) (e^x - 1)/x =lim_(x→0) (sin 3x)/(sin 5x) =d/dx (sin x) =d/dx (eˣ) =d/dx (ln x) =Find d/dx (x² sin x) by product rule:By chain rule, d/dx [sin(x²)] =lim_(x→0) tan(x)/x =lim_(x→∞) (3x² - 5)/(7x² + 2) =l'Hôpital's rule applies to limits of form:Find dy/dx by implicit differentiation for x² + y² = 25:d/dx (xˣ) where x > 0:For f(x) = sin(1/x) for x ≠ 0, what is the limit as x → 0?Differentiation by chain rule for d/dx [ln(sin x)]:lim_(x→0) [(1 + x)^(1/x)] =Higher-order derivative: d²/dx² (x⁴) =f(x) is discontinuous at x = a if which condition fails?For y = x³ - 6x² + 9x, find critical points (where y' = 0):By Taylor expansion around x = 0: e^x = 1 + x + ___: