Question 987100
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When &nbsp;<B>t</B>&nbsp; goes to &nbsp;0, &nbsp;the function {{{sqrt(1+t)-1}}} &nbsp;is equivalent to &nbsp;{{{1/2}}}{{{t}}} &nbsp;and the function &nbsp;{{{sqrt(1-t)-1}}} &nbsp;is equivalent to &nbsp;{{{-1/2}}}{{{t}}}, 

so the difference &nbsp;{{{sqrt(1+t)}}} - {{{sqrt(1-t)}}} &nbsp;is equivalent to 


{{{1/2}}}{{{t)}}} - (-{{{1/2}}}{{{t}}}) = {{{t}}}. 


Therefore, &nbsp;the limit of &nbsp;{{{(sqrt(1+t) - sqrt(1-t))/t}}} &nbsp;at &nbsp;<B>t</B>&nbsp;-->0 &nbsp;is &nbsp;1.