Please show work. Consider the following three functions from R2 to R: Show that
ID: 3343670 • Letter: P
Question
Please show work.
Consider the following three functions from R2 to R: Show that lim(x,yi)rightarrow(0.0) f(x, y) exists, but neither of limxrightarrow0 limyrightarrow0 f(x, y) nor limyrightarrow0 limxrightarrow0) exist. (Recall that by these last two notations we mean just two one-dimensional limits, treating the other variable as a constant in each case.) Show that limxrightarrow0 limyrightarrow0 g(x, y) exists, but neither of lim(x.y)rightarrow0 (o,o) g(x, y) nor limyrightarrow0 limxrightarrow0 g(x, y) exist. Show that limyrightarrow0 limxrightarrow0 h(x, y) exists, but neither of lim(x, y)rightarrow(0, 0) h(x, y) nor limxrightarrow0 limyrightarrow0 h(x, y) exist. This gives cases where all three of these limits give a different answer, so taking limits of each variable independently is quite a different process than a two -dimensional limit! You should feel free to use the facts from lecture that lim(x y )rightarrow(0,0) xy/x2+y2 does not exist and that limxrightarrowo sin 1/x does not exist.Explanation / Answer
1)
this is so because sinx for x in(-infi,infi) is between(-1,1). and we know that 0*(finite no.)=0
but
does not exist, we have to put y as a constant first and then find limit, but we do not know what sin(infi) is
2)
(undefined) does not exist.... here, to prove this, we put y=mx, where m can take any real value. now, if the limit is to exist, the final expression should be independent of m, as m can be anything.... but we get m/(m^2+1) instead.... hence lim doesnt exist
now when we first put y=0, the sin expression becomes 0 as well as the first exp. also becomes 0.
when we put x=0 first, the second expression is undefined
3)it is similar to 2)
lim_{(x,y) ightarrow (0,0)}xsin(1/y)+ysin(1/x)=0