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Evaluate the limits! PLEASE HELP?

0 votes

I cannot solve these limits.. pls help! If the limit does not exist please explain why

b) lim (as x-> pi from the right) (inverse tan(1/(x-pi)))/(pi-x)

d) lim (as x -> 0 from the right) ((sqrt (x+sinx)) lnx)

g) lim ( as x-> 1 from the left) (inverse cos(x))/(1-x) )

asked Apr 20, 2013 in PRECALCULUS by futai Scholar

4 Answers

0 votes

g)

italic lim with x rightwards arrow 1 to the power of minus below fraction numerator italic cos to the power of minus 1 end exponent open parentheses x close parentheses over denominator 1 minus x end fraction equal fraction numerator italic cos to the power of minus 1 end exponent open parentheses 1 close parentheses over denominator 1 minus 1 end fraction           (Substitute x = 1)

space space space space space space space space space space space space space space space space space space space space space space space space equal 0 over 0

space space space space space space space space space space space space space space space space space space space space space space space space equal italic lim with x rightwards arrow 1 to the power of minus below fraction numerator begin display style fraction numerator minus 1 over denominator square root of 1 minus x to the power of 2 end root end fraction end style over denominator 0 minus 1 end fraction     (Apply L'Hospital's rule : italic lim with x rightwards arrow 0 below space fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equal fraction numerator f open parentheses 0 close parentheses over denominator g open parentheses 0 close parentheses end fraction equal 0 over 0 t h e n italic lim with x rightwards arrow 0 below space fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equal italic lim with x rightwards arrow 0 below space fraction numerator f apostrophe open parentheses x close parentheses over denominator g apostrophe open parentheses x close parentheses end fraction)

space space space space space space space space space space space space space space space space space space space space space space space space equal italic lim with x rightwards arrow 1 to the power of minus below fraction numerator 1 over denominator square root of 1 minus x to the power of 2 end root end fraction      

space space space space space space space space space space space space space space space space space space space space space space space space equal fraction numerator 1 over denominator square root of 1 minus 1 to the power of 2 end root end fraction               (Substitute x = 1)

space space space space space space space space space space space space space space space space space space space space space space space space equal infinity                         (Simplify)

answered Apr 25, 2013 by diane Scholar
0 votes

b)

italic lim with x rightwards arrow pi to the power of plus below fraction numerator italic tan to the power of minus 1 end exponent open parentheses begin display style fraction numerator 1 over denominator x minus pi end fraction end style close parentheses over denominator pi minus straight x end fraction equal fraction numerator italic tan to the power of minus 1 end exponent open parentheses begin display style fraction numerator 1 over denominator pi minus pi end fraction end style close parentheses over denominator pi minus pi end fraction    (Substitute x equal pi)

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal fraction numerator italic tan to the power of minus 1 end exponent open parentheses begin display style 1 over 0 end style close parentheses over denominator 0 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal fraction numerator italic tan to the power of minus 1 end exponent open parentheses infinity close parentheses over denominator 0 end fraction     (italic tan open parentheses pi over 2 close parentheses equal infinity )

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal fraction numerator begin display style pi over 2 end style over denominator 0 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal pi over 2 cross times 1 over 0

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal infinity                    (1 over 0 equal infinity )

 

answered Apr 26, 2013 by diane Scholar
0 votes

b)

solution to the problem is 

answered May 14, 2013 by Naren Answers Apprentice
0 votes

d)

solution to the problem is

answered May 14, 2013 by Naren Answers Apprentice

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