Evaluate? #lim/(xrarr0)#(#1/(tsqrt(1+t))-1/t#)

1 Answer
Parabola
Feb 4, 2018

#-1/2#

Explanation:

I am assuming that we have:
#lim_(t->0)(1/(tsqrt(1+t))-1/t)#

Since substituting 0 in the place of #t# gives #1/0-1/0#, let's simplify the function.

We find the LCM between #tsqrt(1+t)# and #t#

We see that it is #tsqrt(1+t)#

We now rewrite our function.
#1/(tsqrt(1+t))-1/t=>1/(tsqrt(1+t))-1/t*(sqrt(1+t))/(sqrt(1+t))#

=>#1/(tsqrt(1+t))-(sqrt(1+t))/(tsqrt(1+t))#

=>#(1-sqrt(1+t))/(tsqrt(1+t))#

Let's try to rationalize the numerator.

=>#(1-sqrt(1+t))/(tsqrt(1+t))*(1+sqrt(1+t))/(1+sqrt(1+t))#

=>#(1-(1+t))/(tsqrt(1+t)+t(1+t))#

=>#(-t)/(tsqrt(1+t)+t+t^2)#

=>#(-t)/(t(sqrt(1+t)+1+t))# We can now cross out the #t#'s.

=>#(-1)/(sqrt(1+t)+1+t)#

We can now plug 0 in the place of #t# to find our limit.

=>#lim_(t->0)(1/(tsqrt(1+t))-1/t)=(-1)/(sqrt(1+0)+1+0)#

=>#lim_(t->0)(1/(tsqrt(1+t))-1/t)=(-1)/(sqrt(1)+1)#

=>#lim_(t->0)(1/(tsqrt(1+t))-1/t)=(-1)/(2)#

=>#lim_(t->0)(1/(tsqrt(1+t))-1/t)=-1/2#

That is our answer!

This works because there was a hole in the graph of our original function. Our new one simply traces over that point, making that point defined. If the function was, say, asymptotically discontinuous (and undefined) at a given point, then our trick would not be able to work.

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