# Thread: solve for limit with square root in numerator and denominator

1. ## solve for limit with square root in numerator and denominator

How would I solve a problem like this
lim (sqrt(3x+3)-3)/(sqrt(10x+5)-5)
x->2
obviously the substitution method doesnt work so I was wondering if someone can help me with this one.

2. Originally Posted by macro
How would I solve a problem like this
lim (sqrt(3x+3)-3)/(sqrt(10x+5)-5)
x->2
obviously the substitution method doesnt work so I was wondering if someone can help me with this one.
multiply the numerator and denominator of the given fraction by (sqrt(10x+5)-5), and you will have

(sqrt(3x+3)-3)*(sqrt(10x+5)-5)
lim -----------------------------------
x->2 (sqrt(10x+5)-5)*(sqrt(10x+5)-5)

NOTE that the denominator above is in the form of (a-b)(a-b), which is equal to a^2-2ab+b^2

(sqrt(3x+3)-3)*(sqrt(10x+5)-5)
= lim --------------------------------------
x->2 (10x+5) - 2*5*(sqrt(10x+5)-5) + 25

calculate the limit as x-> 2, you should get 0.

3. It ends up with a zero in the denominator

4. Originally Posted by macro
How would I solve a problem like this
lim (sqrt(3x+3)-3)/(sqrt(10x+5)-5)
x->2
obviously the substitution method doesnt work so I was wondering if someone can help me with this one.
Hello.

$\lim_{x\to 2} \frac{ \sqrt{3x+3}-3 } { \sqrt{10x+5}-5 } = \sqrt{\frac{3}{5}} \, \lim_{x \to 2 } \frac{ \sqrt{x+1} - \sqrt{3} } { \sqrt{2x+1} - \sqrt{5} }$

Start by multiplying the function by $\frac{\sqrt{x+1}+\sqrt{3}}{\sqrt{x+1}+\sqrt{3}}$.

5. Originally Posted by macro
It ends up with a zero in the denominator
How does it end up with a zero in the denominator?

(sqrt(3x+3)-3)*(sqrt(10x+5)-5)
= lim --------------------------------------
x->2 (10x+5) - 2*5*(sqrt(10x+5)-5) + 25

The denominator will be:

(20+5) - 10*(sqrt(25)-5) +25

= 25 -10*0 +25
= 50

???
The numerator should be a 0 here, and the overall function goes to 0 as x tends to 2!

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# limit of function having three under root function in both nominator and denominator

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