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# Use Definition 8 to prove that $\displaystyle \lim_{x \to -\infty} \frac{1}{x} = 0$.

## For $x < 0,|1 / x-0|=-1 / x$. If $c > 0$ is given, then $-1 / x< c \Leftrightarrow x < -1 / c$$\text { Take } N=-1 / c . \text { Then } x < N \Rightarrow x < -1 / c \Rightarrow|(1 / x)-0|=-1 / x < \varepsilon, \text { so } \lim _{x \rightarrow-\infty}(1 / x)=0$

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This is problem number seventy seven of this tour. Calculus, Eighth edition, Section two point six. Use definition eight to prove that the limit is X approaches. Negative. Infinity. Ah, the function wanna Rex is equal to zero and definite definition eight states that if we have x restricted to pee less than a number. And this should Parenti, that our difference between our function on our limit, thefix and home remains within a certain accuracy. Absalom. So as long as we could do that for any choice of expert Absalom and corresponding and or vice versa, then we have proven that this limit does indeed exist in the sequel two zero. So goingto start with the second part the function. The cervix, which is one of Rex my nest element value, which is zero. Listen, it's invaluable. Solin, leave this with absolutely wanna Rex. Listen, Hasan, I don't want to be careful here. We cannot just remove the absolutely signs since X is in the negative domain. So X is less than we're in a restrict X to only being less than zero as we're approaching infinity. If that is true, then this that equivalent of this absolute value function is negative. One Rex. Less than absolute. We can take the reciprocal and my negative one leaving us with exes. Liston Negative, Absalon on negative one over Absalom and and comparing this results with this condition we can choose and to be equal to negative one over Epsilon. This is one example and this shows that we can indeed find a restriction for X and such for any given Absalon on this case, we restricted apps onto me probably greater than zero. But in this case, if we let say chews up song to be point one than our end corresponding and would be negative one over point one or negative ten. Ah, thank you Could continue to that and through definition eight. Because this is true because all of this is consistent than therefore this limit as expressing an infinity of one of Rex equals zero is true.