What is the limit lim(x tends to 0) (3x^2 - 4x + 7)/x

Expert Answers
embizze eNotes educator| Certified Educator

Find the limit as x tends to zero of (3x^2-4x+7)/x:

Dividing through by x we get the limit as x tends to zero of 3x-4+7/x.

The term 7/x grows without bound as x tends to zero.


The limit as x tends to zero of (3x^2-4x+7)/x does not exist.


We can describe the way the limit fails to exist. We say that the limit as x approaches zero from the right is positive infinity. Infinity is not a real number, so we are not saying that at zero the function is infinite. In this case, the function is not even defined at zero. We are saying that as x tends to zero from the right, the function grows without bound.

As x approaches zero from the left the function decreases without bound, or it is said to go to negative infinity.

Since the left and right "limits" do not agree, we say that there is no limit as x approaches zero.

The graph:

gsarora17 eNotes educator| Certified Educator

A function will have a limiting value only if its right hand limit equals its left hand limit. Let's tabulate the values for both the cases.

x = 0.01 f(x)= 696.03

x= 0.001 f(x) = 6996.003

x= 0.0001 f(x) = 69996.003

x= -0.01  f(x) = -704.03

x= -0.001 f(x) = -7004.003

x= -0.0001 f(x)= -70004.0003

So from the above observations , one sided limits are going to opposite places and hence the limit does not exist as x approaches zero.

kspcr111 | Student

Given to find the limits of 

`lim _(x->0) ((3x^2 - 4x + 7)/x)`


`lim _(x->0) ((3x^2 - 4x + 7)/x)`

The condition for limit does exist or not is 

`lim _(x->0^-)((3x^2 - 4x + 7)/x) =lim _(x->0^+)((3x^2 - 4x + 7)/x)`

so , let us calculate step wise

`lim _(x->0^-)((3x^2 - 4x + 7)/x)`

=> `lim _(x->0^-)((3x - 4 + 7/x))`

=> `lim _(x->0^-) (3x) - lim _(x->0^-) (4) + lim _(x->0^-) (7/x)`

as the denominator of (7/x) is a negative quantity approaching 0

=> `0 - 4 +(-oo) = -oo`

and Now,

`lim _(x->0^+)((3x^2 - 4x + 7)/x)`

=>`lim _(x->0^+)((3x - 4 + 7/x))`

=>`lim _(x->0^+) (3x) -lim _(x->0^+) (4) +lim _(x->0^+) (7/x)`

as the denominator of (7/x) is a positive quantity approaching 0

=> `0 - 4 +(+oo) = +oo`

Now as we observe 

`lim _(x->0^-)((3x^2 - 4x + 7)/x) !=lim _(x->0^+)((3x^2 - 4x + 7)/x)`  

so , the limit doesn't exist