Sorry for the weird title, I don't know how to put the equation on the title.
$$\lim_{x\to-\infty}\left(\frac{1-x^3}{x^2+7x}\right)^5$$
Ok I divided inside the parenthesis by $x^2$, but now I am stuck.
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2$\begingroup$ You'd be better off dividing by $x^3$ inside the parantheses. $\endgroup$– Mike EarnestCommented Oct 13, 2014 at 6:16
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$\begingroup$ I got -1/0 for the answer?? $\endgroup$– ElsaCommented Oct 13, 2014 at 6:18
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$\begingroup$ Can I use L'hospital's rule? $\endgroup$– JohnCommented Oct 13, 2014 at 6:23
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$\begingroup$ no we are not allowed to use that rule yet $\endgroup$– ElsaCommented Oct 13, 2014 at 6:34
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2 Answers
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If we divide numerator and denominator in the parentheses by $x^3$ we get
$$\frac{1-x^3}{x^2+7x}=\frac{ 1/x^3 - 1}{1/x + 7/x^2}$$
Now note that as $x\to -\infty$ the numerator approaches $-1$ and the denominator approaches $0$. So overall the fraction diverges. This already shows that the original limit diverges since the exponential function is continuous.
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Since $X^5$ is continuous, it suffice to calculate $$\lim_{x\to-\infty}\left(\frac{1-x^3}{x^2+7x}\right)$$. Then by using L'Hospital's rule twice, you get the answer $\infty$.
Without using LHospital's rule, divide $x^2$ on the numerator and denominator. $$\lim_{x\to-\infty}\left(\frac{\frac{1}{x^2}-x}{1+\frac{7}{x}}\right)$$
As $x\to -\infty$, $\frac{1}{x^2}\to 0, \frac{1}{x}\to 0$, hence numerator $\to \infty$, denominator $\to 1$, so the limit is $\infty$
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$\begingroup$ how do you get the answer without the rule? we are not allowed to use it yet $\endgroup$– ElsaCommented Oct 13, 2014 at 6:34
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$\begingroup$ @Kaaagome I have added more details. $\endgroup$– JohnCommented Oct 13, 2014 at 6:39
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$\begingroup$ Thank you for your reply! I did divide by x^2, but I'm still confused how the numerator becomes infinity? I know that 1/x^2 = 0, but what about -x in the numerator? $\endgroup$– ElsaCommented Oct 13, 2014 at 6:50
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$\begingroup$ @Kaaagome $-x\to \infty$, which means it can be larger that any positive number. Hence the numerator $\to \infty$. I think you may be not clear about the definition of $\lim_{x\to-\infty}f(x)=-\infty$. It means $\for all M<0, \exists N\in \mathbb{R}$,s.t. $\forall x<-N, f(x)<M$. You can use this to check the numerator. $\endgroup$– JohnCommented Oct 13, 2014 at 7:01
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$\begingroup$ ah so you mean like u plugged in negative infinity for x? so infinity/1 = infinity. But if it was like infinity/2, would that still be infinity? $\endgroup$– ElsaCommented Oct 13, 2014 at 7:07