schoolAccedmy
9th10th
11th12th
school

11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)lim⁡x→3(2x+4)\lim_{x \to 3}(2x + 4)limx→3​(2x+4)menu_book•(ii)lim⁡x→1(3x2−2x+4)\lim_{x \to 1}(3x^2 - 2x + 4)limx→1​(3x2−2x+4)menu_book•(iii)lim⁡x→5x2+x+4\lim_{x \to 5}\sqrt{x^2 + x + 4}limx→5​x2+x+4​menu_book•(iv)lim⁡x→2x2+4\lim_{x \to 2}\sqrt{x^2 + 4}limx→2​x2+4​menu_book•(v)lim⁡x→2(x3+1−x2+5)\lim_{x \to 2}(\sqrt{x^3 + 1} - \sqrt{x^2 + 5})limx→2​(x3+1​−x2+5​)menu_book•(vi)lim⁡x→22x3+5x3x−2\lim_{x \to 2}\dfrac{2x^3 + 5x}{3x - 2}limx→2​3x−22x3+5x​
arrow_backBack
school

11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)lim⁡x→3(2x+4)\lim_{x \to 3}(2x + 4)limx→3​(2x+4)menu_book•(ii)lim⁡x→1(3x2−2x+4)\lim_{x \to 1}(3x^2 - 2x + 4)limx→1​(3x2−2x+4)menu_book•(iii)lim⁡x→5x2+x+4\lim_{x \to 5}\sqrt{x^2 + x + 4}limx→5​x2+x+4​menu_book•(iv)lim⁡x→2x2+4\lim_{x \to 2}\sqrt{x^2 + 4}limx→2​x2+4​menu_book•(v)lim⁡x→2(x3+1−x2+5)\lim_{x \to 2}(\sqrt{x^3 + 1} - \sqrt{x^2 + 5})limx→2​(x3+1​−x2+5​)menu_book•(vi)lim⁡x→22x3+5x3x−2\lim_{x \to 2}\dfrac{2x^3 + 5x}{3x - 2}limx→2​3x−22x3+5x​
arrow_backBack
Accedmychevron_right11thchevron_rightmathchevron_rightLimit And Continuitychevron_rightExercise 12.1
arrow_back
Previous Question
Question 1
Next Question
Question 3
arrow_forward
On this page
No headings yet