Homework SourseUse the squeeze theorem to show that lim sqrtx3 x2 sinpix
Use the squeeze theorem to show that.
lim sqrt[x^3+ x^2] * sin(pi/x) = 0
x->0
lim sqrt[x^3+ x^2] * sin(pi/x) = 0
x->0
Solution
sin(/x) deos not exists at x=0 but the def of sine x is
-1<= sin(pi/x) <= 1
hence
-sqrt[x^3+ x^2] < = f(x->0) < = sqrt[x^3+ x^2]
by the theorem of squeezing or sandwitch
the f(x) at 0 is 0
because both left limit and right limit are 0
sqrt[x^3+ x^2] = 0 for x= 0
so limit becomes 0<= f(0) <= 0
hence f(x-> 0 ) = 0
![Use the squeeze theorem to show that. lim sqrt[x^3+ x^2] * sin(pi/x) = 0 x->0Solution sin(/x) deos not exists at x=0 but the def of sine x is -1<= sin(pi/](/WebImages/43/use-the-squeeze-theorem-to-show-that-lim-sqrtx3-x2-sinpix-1133442-1761606061-0.webp "Use the squeeze theorem to show that. lim sqrt[x^3+ x^2] * sin(pi/x) = 0 x->0Solution sin(/x) deos not exists at x=0 but the def of sine x is -1<= sin(pi/")
