please, help me out to this limit problem
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

I'm assuming the function is (sqrt(x+sqrt(x))/(x^2-16). Two things to note: the function can't be defined for x=4, because the denominator would be zero; and the function can't be defined for any negative values of x, because we have a term sqrt(x) and we can't take the real square root of a negative number (I'm assuming we're not dealing with imaginary numbers).

We have two scenarios for values of x:

  1. 0<=x<4
  2. x>4

In (1) the function becomes negative, and in (2) it becomes positive. x=4 is an asymptote. As x approaches 4 from the "left", that is, just less than 4, the value of the function approaches a large negative multiple of sqrt(6). As x approaches 4 from the "right", just greater than 4, the value of the function approaches a large positive multiple of sqrt(6). (Incidentally, at x=0, the function is zero, and as x gets very large and positive the function approaches zero.) I hope this helps.

by Top Rated User (1.2m points)
reshown by

Related questions

2 answers
1 answer
asked Sep 29, 2021 in Calculus Answers by anonymous | 956 views
1 answer
0 answers
asked Oct 31, 2018 in Calculus Answers by Bibin | 564 views
1 answer
asked Aug 13, 2018 in Calculus Answers by claire | 1.4k views
1 answer
asked Jun 5, 2018 in Calculus Answers by anonymous | 497 views
1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,516 questions
100,344 answers
2,420 comments
763,911 users