Sunday, 11 June 2017

calculus - Two square roots in an indeterminate




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
limx5x2+530(x5)




This is where I've gotten
limx5x2+530(x5)x2+5+30x2+5+30=x2+530(x5)(x2+5+30)


Answer



HINT     Factoring both of  x2+52302 = x252  by difference of squares yields



x2+530x5 = x+5x2+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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