MATH-UA 211 Lecture 7: Limits and Continuity

• School: New York University
• Department: Math
• Course: MATH-UA 211
• Professor: Majmudar

Document Summary

Limits: lim x 3 x2+4 x 3 x+1 lim x 2 (direct substitution since there is no jump/hole), lim= 13 lim= (-1)/(3) 1/6 (cancel out the (x: and then substitute 2 into the equation. 16+8h+h2 16 h (8h+h2) h: factor out h in the numerator h(8+h) and simplify- limit=8. X x xif x 0 xif x<0. X o: evaluate one sided limits: =1 x 0+ x x lim x 0 x x lim . A junction is continuous at a point x=a if lim (x a) f(x)=f(a) Limit exists if: f(a) is defined, lim (x a) f(x) exists, lim (x a) f(x)= f(a) Similar to law of limits: if f and g are continuous at x=a, then so are f+g , f-g, c*f, and f*g, f/g given that g 0. Not continuous at x=2 x x2 3x 4 -not continuous at x=4 and x=-1.