It is proved by definition that when x tends to 1, LIM (3x + 2) = 5

It is proved by definition that when x tends to 1, LIM (3x + 2) = 5

For arbitrary ε> 0
take δ=ε/ three
When X-1|< δ When,
|3x+2 -5|=3|x-1|<3* ε/ 3= ε
So LIM (X -- > 1) (3x + 2) = 5

LIM (x trend - 1) (2x ^ 2-x-3) / (x + 1) = - 5 proved by definition

LIM (x trend - 1) (2x ^ 2-x-3) / (x + 1) = - 5 prove it by definition, thank you
Man, isn't that what this question is about?
LIM (x trend - 1) (2x ^ 2-x-3) / (x + 1) = LIM (x trend - 1) (2x-3) (x + 1) / (x + 1) = LIM (x trend - 1) (2x-3)
=-3 + 2 * LIM (x trend - 1) x = - 5

Limit of LIM (x → π) SiN x / X - π

Let a = x - π
Original formula = LIM (a → 0) sin (π + a) / A
=lim(a→0)(-sina)/a
=-1

Find the limit of LIM (x → 0) [(x-sinx) ÷ (x + SiNx)] Give me some steps to solve the problem. Thanks

0...... take the derivative of (1-cosx) / (1 + cosx) for the numerator and denominator respectively, and substitute x equal to zero to obtain the value of 0

Find the limit LIM (x → + ∞) (x-sinx) / x ^ 3

The key to solving the problem: the product of inverse substitution, infinitesimal and bounded variables is still infinitesimal
This question is a little strange. It should be asked that the limit of the original proposition is meaningful when x tends to 0

When x → 0, find the limit of LIM X / SiNx? But my answer is 0, X is infinitesimal, and 1 / SiNx is a bounded function, so the answer is 0. I don't know what's wrong,

Isn't this one of the two important limits? The limit is 1