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\[\lim_{x \rightarrow 121}\frac{ \sqrt{x}-11 }{ x-121 }\]

can we factor the bottom?

if x was squared then yes but it's not.

oh but it is ... let a^2 = x, then a = sqrt(x)

wait I don't understand that, can you explain it further?

at x=11, we have 0/0 right?

do you mean at x=121?

how would you go about replacing it with something that looks "squarey"?

your notes might say; divide it al by the highest power of x tho

yes

and since sqrt(x) = a
sqrt(121) = 11 and we get your coveted 1/22

Thank you!

factoring seems to be the best route to me ... and youre welcome again :)