How do you find the limit of #(1/(3+x)-(1/3))/(x)# as #x-0#?

Question

How do you find the limit of #(1/(3+x)-(1/3))/(x)# as #x->0#?

Aurora Alvarado · Accepted Answer

#lim_(xrarr0)(1/(3+x)-1/3)/x=-1/9#

#lim_(xrarr0)(1/(3+x)-1/3)/x#

We can clear this limit of fractions by multiplying as follows:

#=lim_(xrarr0)((1/(3+x)-1/3)(3)(3+x))/(x(3)(3+x))#

Multiplying through in the numerator gives:

#=lim_(xrarr0)((3(3+x))/(3+x)-(3(3+x))/3)/(3x(x+3))#

Canceling:

#=lim_(xrarr0)(3-(3+x))/(3x(x+3))#

#=lim_(xrarr0)(3-3-x)/(3x(x+3))#

#=lim_(xrarr0)(-x)/(3x(x+3))#

#=lim_(xrarr0)(-1)/(3(x+3))#

Now we can evaluate the limit, since there is no longer an issue for #xrarr0#.

#=(-1)/(3(0+3))#

#=-1/9#

Evelyn Agard · Answer

undefined To find the limit of the expression (1/(3+x)-(1/3))/(x) as x approaches 0, we can simplify the expression first.

First, we need to find a common denominator for the two fractions in the numerator. The common denominator is 3(3+x), which gives us:

[(3(3+x) - (3+x))/(3(3+x))]/x

Simplifying further, we have:

[(9+3x - 3 - x)/(3(3+x))]/x

Combining like terms, we get:

(6+2x)/(3(3+x))/x

Now, we can simplify the expression by canceling out common factors. Canceling out the x in the numerator and denominator, we have:

(6+2x)/(3(3+x))

Next, we can substitute x=0 into the simplified expression:

(6+2(0))/(3(3+0))

Simplifying further, we get:

6/9

Finally, we can simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 3:

2/3

Therefore, the limit of the expression (1/(3+x)-(1/3))/(x) as x approaches 0 is 2/3.

How do you find the limit of #(1/(3+x)-(1/3))/(x)# as #x-&gt;0#?