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?

Paisley Johnson · Accepted Answer

Assuming you meant:

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

Multiply the fraction by the inner denominators:

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

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

Now we can evaluate the limit:

#=(-1)/(3(0+3))=-1/9#

Charles Arocha · 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, let's simplify the expression [1/(3+X)]- (1/3) ÷ X:

[1/(3+X)]- (1/3) ÷ X = [1/(3+X)] - (1/3) * (1/X)

Next, let's find a common denominator for the two fractions:

[1/(3+X)] - (1/3) * (1/X) = [1/(3+X)] - (1/3) * (X/(X))

Simplifying further:

[1/(3+X)] - (1/3) * (X/(X)) = [1/(3+X)] - (X/3X)

Now, let's combine the fractions:

[1/(3+X)] - (X/3X) = [1/(3+X)] - (X/3X) * (3/3)

Simplifying again:

[1/(3+X)] - (X/3X) * (3/3) = [1/(3+X)] - (3X/9X)

Combining the fractions:

[1/(3+X)] - (3X/9X) = [1/(3+X)] - (1/3)

Now, we can find the limit as x approaches 0:

lim(x->0) [1/(3+X)] - (1/3) = 1/3 - 1/3 = 0

Therefore, the limit of the expression [1/(3+X)]- (1/3) ÷ X as x approaches 0 is 0.

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