AboutPaul Klarreich Expertise All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions.
I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.
Experience I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.
Using the formal epsilon − delta definition of derivative, show that
lim xapprochaes 0 x^2sin(x)=0
Answer Questioner: waqas
Country: Australia
Category: Calculus
Private: No
Subject: Epsilon Delta problems
Question: plz solve this for me
Using the formal epsilon − delta definition of derivative, show that
lim xapprochaes 0 x^2sin(x)=0
..................................................
Hi, Wagas,
You are not using the vocabulary correctly. You should either write:
Using the formal epsilon − delta definition of limit
or
Using the formal definition of derivative as a limit.
Misusing (and therefore misunderstanding) the vocabulary is a good way to fail mathematics.
............................................
I shall assume you mean:
Using the formal epsilon − delta definition of limit
show that lim [x -> 0] x^2sin(x)=0
You want to show that for any e > 0 [e is epsilon, d is delta]
you can find d > 0 such that whenever |x - 0| < d,
| x^2sin(x) - 0 | < e
Start with
| x^2sin(x) - 0 | = | x^2sin(x) |
| x^2sin(x) | = | x^2 | | sin(x) |
| x^2 | | sin(x) | <= | x^2 | (1) = | x^2 | = x^2, since x^2 is never negative.
