Square roots are always causing trouble for me - especially finding the limit of a function when it's in the indeterminate form.
I got this far, but I don't know what to do next or even if it's right. Trying to go any further gives me my original problem.
This is the initial problem
limx→5√x2+5−√30(x−5)
This is where I've gotten
limx→5√x2+5−√30(x−5)∗√x2+5+√30√x2+5+√30=x2+5−30(x−5)(√x2+5+√30)
Answer
HINT Factoring both of √x2+52−√302 = x2−52 by difference of squares yields
√x2+5−√30x−5 = x+5√x2+5+√30
Alternatively, if you're familiar with derivatives, note that the limit is f′(5) for f(x)=√x2+5.
Many limits can be simply calculated by recognizing them as instances of first derivatives and then calculating the derivatively rotely using known derivative rules. You can find a handful of examples in my prior posts starting here. Later in your course you will most likely encounter a generalization of this observation known as l'Hôpital's Rule.
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