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

1 Answer
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.