1could somebody elaborate???
There's the general rukle for limits evaluation .
lim x→a f(g(x)) = f( lim x→a g(x) ).
but I feel , there must be someother condition . cos of this
lim x--> 0 [x3] ≠[lim x --> 0 x3] .
help karo yaar.
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7 Answers
341that condition is continuity. This follows from Heine's definition of continuity
62you need continuity of f i think...
you probably can do without the continuity of g...
1@prophet . I'm getting a bit confused da.
http://en.wikipedia.org/wiki/Continuous_function#Heine_definition_of_continuity
At wiki it says for fog , f is continuous if g is tending to a limit.
but here g does tend to a limit but f isn't continouus no???
koi help karo . maybe I'm not understanding it properly( what's given at wiki) . madad karo koi . check out the link.
1@srinath,,, its clearly given in fiitjee's RSM dat:
let lim x->a f(x)=m and lim x->a g(x)=n
lim x->a f(g(x)) = f(lim x->a g(x)) = f(n)
if and only if f(x) is continuos at x=n.
did u get it?
the function f(x) has to be continuous at the limiting value of g(x->a)
for the ex u gave,
g(x)= x3 , f(x)=[x].
for limx->a g(x)=a3
but at x=a3, f(x) is not continuous... (provided 'a' is integre)
so the formula is not valid...
341Firstly, let's see what is the definition of limit as given by Heine
Lim x→ a F(x) exists iff for every sequence x1,x2,...,xn,... (we can write {xn} for short) converging to a, the sequence {F(xn} converges to a limit l.
If F(x) is continuous this further means that the sequence {F(xn)} will converge to F(a)
So, now let us look at the question Lim x→ a f(g(x))
So first we must look at all sequences {xn)} converging to a.
This in turn gives rise to a subsequence,{g(xn)}
If the given limit exists, this subsequence will converge to some limit l
Now further if f(x) is continuous, we must have
that the {f(g(xn))} will converge to f(l)
i.e. to f(Lim x→ a g(x))
Hence, iff f(x) is continuous and Lim x→ a g(x) exists, then Lim x→ a f(g(x)) = f(Lim x→ a g(x))
341Srinath, if you are not still satisfied, please do post your doubt. I was not online much yday, so I couldnt reply in time