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Here are my limits:

$$\lim_{x\to -2^+}\frac{-4}{x+2}$$ right hand limit $$\lim_{x\to -2^-}\frac{-4}{x+2}$$ left hand limit

My question: The first limit evaluates to -infinity, but why? As we approach -2 from the right we get closer and closer to -2. So we pass by +1, 0, enter in on the negative numbers like -1 and start getting close to -2 right? In fact we are interested what happens infinitely close to -2, but only on the right hand side for this limit. Are not the values of x close to -2 on the right hand side still negative numbers? Eg, -1.8, -1.999999 ect... So why is the limit equal to -infinity? -4/-1.99999999999 is a large positive quantity right? Whats wrong with my thinking?

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3 Answers 3

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You are right, the numbers are negative, but if $x$ approaches $-2$ from the right, $x+2$ is always positive. Hence, the limit is $-\infty$ because $-4$ is negative.

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As we approach $-2$ from the right we are looking at values of $x$ that are greater than $-2$, and as such $x+2>0$ and the fraction is negative. This is why it's negative infinity from the right. A similar argument explains why it's positive infinity from the left.

Be careful with your calculations. If you choose $x=-1.999$ the denominator is $0.001$, a positive number. I find it helpful to look at the graph in these situations. https://www.wolframalpha.com/input/?i=-4%2F(x%2B2)

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Nothing is wrong with your thinking! Look at the graph below.

enter image description here

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