Question 178429
Evaluate the limit of each indeterminate quotient:
<pre><font size = 4 color = "indigo"><b>
{{{matrix(3,2, 
             "",    "",      
            lim,  (sqrt(5-x)-sqrt(3+x))/(x-1),
            "x->1",    "" )}}}

The numerator has the square roots, so we 
multiply top and bottom by the conjugate of the
numerator, which is {{{(sqrt(5-x)+sqrt(3+x))}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  (  (sqrt(5-x)-sqrt(3+x))(sqrt(5-x)+sqrt(3+x))  )/((x-1)(sqrt(5-x)+sqrt(3+x))),
            "x->1",    "" )}}}

If you "FOIL" out the top the middle two terms
cancel and we get:

{{{matrix(3,2, 
             "",    "",      
            lim,  ((5-x)-(3+x))/((x-1)(sqrt(5-x)+sqrt(3+x))),
            "x->1",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  (5-x-3-x)/((x-1)(sqrt(5-x)+sqrt(3+x))),
            "x->1",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  (2-2x)/((x-1)(sqrt(5-x)+sqrt(3+x))),
            "x->1",    "" )}}}

Factor {{{-2}}} out of the numerator:


{{{matrix(3,2, 
             "",    "",      
            lim,  (-2(-1+x))/((x-1)(sqrt(5-x)+sqrt(3+x))),
            "x->1",    "" )}}}

Write the expression in the parentheses on top in 
descending order:

{{{matrix(3,2, 
             "",    "",      
            lim,  (-2(x-1))/((x-1)(sqrt(5-x)+sqrt(3+x))),
            "x->1",    "" )}}}

Cancel the {{{(x-1)}}}'s:

{{{matrix(3,2, 
             "",    "",      
            lim,  (-2(cross(x-1)))/((cross(x-1))(sqrt(5-x)+sqrt(3+x))),
            "x->1",    "" )}}}

{{{matrix(3,12,"","","","","","","","","","", 
"",    "",      
lim,(-2)/(sqrt(5-x)+sqrt(3+x)),"=",(-2)/(sqrt(5-1)+sqrt(3+1)),"=",(-2)/(sqrt(4)+sqrt(4)),"=",(-2)/(2+2),"=",(-2)/(4),"=",-1/2,  
            "x->1",    "","","","","","","","","","","" )}}}


------------------------------------------------

{{{matrix(3,2, 
             "",    "",      
            lim,  (2-sqrt(x))/(3-sqrt(2x+1)),
            "x->4",    "" )}}}

Both the numerator and the denominator have 
square roots, so we multiply top and bottom by 
the conjugates of both numerator and denominator,
which is {{{(2+sqrt(x))(3+sqrt(2x+1))}}}.

We'll reverse the order of those factors when we
multiply the bottom by it so the conjugates will 
be next to what they're the conjugates of:

{{{matrix(3,2, 
             "",    "",      
            lim,  ( (2-sqrt(x))(2+sqrt(x))(3+sqrt(2x+1)) )/((3-sqrt(2x+1))(3+sqrt(2x+1))(2+sqrt(x))),
            "x->4",    "" )}}}

"FOIL" out the first two terms in the top, and the 
two middle terms will cancel.  Do the same in the
bottom:

{{{matrix(3,2, 
             "",    "",      
            lim,  ( (4-x)(3+sqrt(2x+1)) )/((9-(2x+1))(2+sqrt(x))),
            "x->4",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  ( (4-x)(3+sqrt(2x+1)) )/((9-2x-1)(2+sqrt(x))),
            "x->4",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  ( (4-x)(3+sqrt(2x+1)) )/((8-2x)(2+sqrt(x))),
            "x->4",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  ( (4-x)(3+sqrt(2x+1)) )/(2(4-x)(2+sqrt(x))),
            "x->4",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  ( (cross(4-x))(3+sqrt(2x+1)) )/(2(cross(4-x))(2+sqrt(x))),
            "x->4",    "" )}}}

{{{matrix(3,14, 
"",    "","","","","","","","","","","","","",      
lim,   (3+sqrt(2x+1)) /(2(2+sqrt(x))),"=",(3+sqrt(2*4+1)) /(2(2+sqrt(4))),"=",(3+sqrt(8+1))/(2(2+2)),"=",(3+sqrt(9))/(2(4)),"=",(3+3)/8,"=",6/8,"=",3/4,
 "x->4",    "","","","","","","","","","","","","" )}}}

----------------------------------------

{{{matrix(3,2, 
             "",    "",      
            lim,  (2^(2x)-2^x)/(2^x-1),
            "x->0",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  ((2^x)^2-2^x)/(2^x-1),
            "x->0",    "" )}}}

Factor {{{(2^x)}}} out of the top:

{{{matrix(3,2, 
             "",    "",      
            lim,  ((2^x)(2^x-1))/(2^x-1),
            "x->0",    "" )}}}

{{{matrix(3,2, 
             "",    "",      
            lim,  ((2^x)(cross(2^x-1)))/(cross(2^x-1)),
            "x->0",    "" )}}}

{{{matrix(3,6, 
             "",    "", "","","","",     
            lim,  2^x, "=", 2^0, "=",1, 
            "x->0",    "","","","","" )}}}

Edwin</pre>