# Evaluate the limit [sqrt(x+6)-4]/(x-10) x-->10

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### 4 Answers

If we calculate [sqrt(x+6)-4]/(x-10) for x=10 we get

[sqrt(10+6)-4]/(10-10) = [sqrt(16)-4]/(10-10) = [(4-4)/(10-10)]=0/0 which is not defined.

Therefore to find the expression [sqrt(x+6)-4]/(x-10) for the limit x-->10, we have to simplify it.

[sqrt(x+6)-4]/(x-10)= [sqrt(x+6)-4]/(x+6-10-6)

=[sqrt(x+6)-4]/[x+6-16]

=[sqrt(x+6)-4]/[sqrt(x+6)-sqrt 16]*[sqrt(x+6)+sqrt 16]

=[sqrt(x+6)-4]/[{sqrt(x+6)-4}*{sqrt(x+6)+4}]

=1/[sqrt(x+6)+4]

Now substituting x=10 we get

1/[sqrt(10+6)+4]

=1/ [sqrt 16+4]

=1/ [4+4]

=1/8

**Therefore [sqrt(x+6)-4]/(x-10) for limit x-->10 is 1/8.**

To find the lt {sqrt(x+4)-4}/(x-10) as x--> 10.

Solution:

f(x) = Lt {sqrt(x+6)-4]/(x-10)} as x--> 0.

Put x = 10 , Then f(10) Lt{ sqrt(10+6)-4}/(10-10) = 0/0 form.

Since both numerator and denomonator go zero, we consider rationalising the numerator and divide by common factors of numerator and denominator.

f(x)= {sqrt(x+6)-4)}(sqrt(x+6)+4}/(x-10){sqrt(x+6)+4}

f(x)= {(x+6) - 4^2}/{(x-10)[sqrt(x+6)+4]}

f(x) = (x-10)/{(x-10){sqrt(x+6)+4}

f(x) = 1/{sqrt(x+6)+4}.

Now take limit as x --> 10.

Ltf(x) = lt 1/{sqrt(x+6)+4} = 1/(sqrt(10+6)+4} = 1/(4+4) = 1/8.

In order to evaluate this limit, first we'll verify if it is an indetermination case, and it is, "0/0" case, so, we'll apply L'Hospital rule.

Let's recall L'Hospital rule.

lim [f(x)/g(x)] = lim {[f(x)]' / [g(x)]'} if and only if

lim f(x) = 0 and lim g(x) = 0

or

lim f(x) = inf and lim g(x) = inf

We'll substitute x by 10 and we'll get an indetermination case "0/0".

Let's denote f(x) = sqrt(x+6)-4 and g(x) = x-10

Let's apply L'Hospital rule now:

lim {[sqrt(x+6)-4]/(x-10)}=lim {[sqrt(x+6)-4]'/(x-10)'}

[sqrt(x+6)-4]' = (x+6)'/2*sqrt(x+6) = 1/2*sqrt(x+6)

(x-10)'=1

lim {[sqrt(x+6)-4]'/(x-10)'}= lim[1/2*sqrt(x+6)]=1/2*sqrt(10+6)

**lim {[sqrt(x+6)-4]/(x-10)} = 1/2sqrt16**

**lim {[sqrt(x+6)-4]/(x-10)} = 1/8**

The limit `lim_(x->10) (sqrt(x+6)-4)/(x-10)` has to be determined.

If we substitute x = 10 in the expression `(sqrt(x+6)-4)/(x-10)` we get the indeterminate form `0/0` . Use l'Hospital's rule and take the derivative of the numerator and the denominator.

This gives:

`lim_(x->10) ((1/2)*1/(sqrt(x+6)))/1`

Now replacing x = 10 gives `((1/2)*1/(sqrt(10+6)))/1` = `(1/2)*1/(sqrt 16)` = `(1/2)*(1/4)` = `1/8`

The limit `lim_(x->10) (sqrt(x+6)-4)/(x-10) = 1/8`