I want to investigate the value of limx→0+e1x2sinx. Since the expontial tends really fast to infinity but the sine quite slowly to 0 in comparison I believe the limit to be infinity. But I cannot find I way to prove it. I tried rewriting using the standard limit sinxx as sinxx⋅xe1x2 but I still get an indeterminate form "1⋅0⋅∞".
Subscribe to:
Post Comments (Atom)
real analysis - How to find limhrightarrow0fracsin(ha)h
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
Ok, according to some notes I have, the following is true for a random variable X that can only take on positive values, i.e P(X\int_0^...
-
Self-studying some properties of the exponential-function I came to the question of ways to assign a value to the divergent sum $$s=\sum_{k=...
-
The question said: Use the Euclidean Algorithm to find gcd (1207,569) and write (1207,569) as an integer linear combination of 1207 ...
No comments:
Post a Comment