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I have determined so far that this is equal to $$\lim_{x \to -1} \frac {(x+1)(x+1)}{(x+\sqrt [4] {1})(x-\sqrt [4]{1})(x^2+\sqrt{1})}.$$

However, my numerator becomes $0$ if I substitute the limit. What am I doing wrong?

Travis Willse
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user137452
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  • Why does it matter if the numerator becomes $0$? Currently your denominator becomes $0$ as well, but you can cancel with the numerator to fix this. –  Oct 10 '14 at 01:36
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    The expression in the body of the post isn't equivalent to the one in the title. – Travis Willse Oct 10 '14 at 01:37
  • A $0$ in the numerator is never a problem. It's the $0$ in the denominator that causes issues. – J126 Oct 10 '14 at 01:38
  • Thus, the only time that I need to factor the numerator is when the denominator equals 0, other than that, if the denominator equals a number > 0 I can calculate the given quotient, as long as the denominator is not 0. – user137452 Oct 10 '14 at 01:56

2 Answers2

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The expression whose limit you're taking is a function continuous at $x = -1$, so the limit is just given by evaluating, which you've essentially already done:

$$\lim_{x \to -1} \frac{x^2 + 2x + 1}{x^2 + 4} = \frac{(-1)^2 + 2(-1) + 1}{(-1)^2 + 4} = 0.$$

Travis Willse
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  • I guess the better question is, when is 0 an acceptable answer. for example, lim -->5 $\frac {x^2-6x+5}{x-5}\ $ one factors the numerator to exclude the denominator. Do you understand what I am seeking to understand? – user137452 Oct 10 '14 at 01:41
  • Yes, but if this is your real question, you should ask that instead or, at this point, perhaps as a new question. If you post it I'm sure you'll get useful replies. – Travis Willse Oct 10 '14 at 01:43
  • $0$ is always an acceptable answer if you achieve it by working legitimately. Is there any particular reason you think $0$ is bad/different from other numbers ? – AgentS Oct 10 '14 at 01:45
  • Thus, the only time that I need to factor the numerator is when the denominator equals 0, other than that, if the denominator equals a number > 0 I can calculate the given quotient, as long as the denominator is not 0. – user137452 Oct 10 '14 at 01:57
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The limit of this function is zero because when you substitute -1 in the function the denominator does not come to be zero.When ever the denominator comes to be zero then only we need to simplify the function,or else not.

tantheta
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