How do you evaluate #lim_(x-0) (sqrt(4+x)-2)/(3x)#?

Question

How do you evaluate #lim_(x->0) (sqrt(4+x)-2)/(3x)#?

Luke Alvarez · Accepted Answer

#lim_(x->0) (sqrt(4+x)-2)/(3x) = 1/12#

Apply L'Hospital's Rule :

#lim_(x->0) (sqrt(4+x)-2)/(3x)#

#= lim_(x->0) ((d((4+x)^(1/2)-2))/(dx))/((d(3x))/(dx))#

#=lim_(x->0) (1/2(4+x)^(-1/2))/(3)#

#=lim_(x->0) 1/(6sqrt(4+x))#

#=1/(6sqrt(4))#

#=1/12#

Paisley Johnson · Answer

Given: #lim_(x to 0)(sqrt(4+x)-2)/(3x)#

Because the limit yields the indeterminant form #0/0#, one should use L'Hôpital's Rule. But, because the topic is "Determining Limits Algebraically", I shall assume that the student has not, yet, learned L'Hôpital's Rule and refrain from using it.

Multiply the expression by 1 in the form of #(sqrt(4+x)+2)/(sqrt(4+x)+2)#:

#lim_(x to 0)(sqrt(4+x)-2)/(3x)(sqrt(4+x)+2)/(sqrt(4+x)+2)#

The numerator becomes the difference of two squares:

#lim_(x to 0)((sqrt(4+x))^2-(2)^2)/((3x)(sqrt(4+x)+2))#

Expand the squares:

#lim_(x to 0)(4+x-4)/((3x)(sqrt(4+x)+2))#

Simplify the numerator:

#lim_(x to 0)x/((3x)(sqrt(4+x)+2))#

#x/x# becomes 1:

#lim_(x to 0)1/(3(sqrt(4+x)+2))#

Now, we may evaluate at #x = 0#:

#1/(3(sqrt(4)+2)) =1/12#

This limit is the same as the original expression:

#lim_(x to 0)(sqrt(4+x)-2)/(3x) = 1/12#

Elijah Abram · Answer

undefined To evaluate the limit lim_(x->0) (sqrt(4+x)-2)/(3x), we can use algebraic manipulation and the limit properties. First, we can simplify the expression by multiplying the numerator and denominator by the conjugate of the numerator, which is sqrt(4+x) + 2. This will help us eliminate the square root in the numerator.

After multiplying, we get (sqrt(4+x) - 2)(sqrt(4+x) + 2)/(3x)(sqrt(4+x) + 2). Simplifying further, we have (4+x - 4)/(3x)(sqrt(4+x) + 2). The numerator cancels out, leaving us with x/(3x)(sqrt(4+x) + 2).

Next, we can simplify the expression by canceling out the common factor of x in the numerator and denominator. This gives us 1/(3(sqrt(4+x) + 2)).

Now, we can evaluate the limit as x approaches 0. Plugging in 0 for x, we get 1/(3(sqrt(4+0) + 2)) = 1/(3(2)) = 1/6.

Therefore, the limit of (sqrt(4+x)-2)/(3x) as x approaches 0 is 1/6.

How do you evaluate #lim_(x-&gt;0) (sqrt(4+x)-2)/(3x)#?

How do you evaluate #lim_(x->0) (sqrt(4+x)-2)/(3x)#?