Limits Calculus Problem
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skeeter
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the limit of the rational, second term is the ratio of the leading coefficients, i.e. -1/2
the limit of the first term, [math]\cos{x}[/math] , as [math]x \to -\infty[/math] does not exist (why?)
so, what does that say about the overall limit?
Steven G
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You ask if the limit is infinity (or any other value). Why not just evaluate the expression with x = 100,000,000 (a large number).
That is calculate cos(100,000,000) + 100,000,000/(7-2*100,000,000). What do you get? The limit will be close to that. Now go and try to figure out how to get the correct answer. Remember that cos x will simply fluctuate between -1 and 1 so keep that in mind when you calculate cos(100,000,000).
That is calculate cos(100,000,000) + 100,000,000/(7-2*100,000,000). What do you get? The limit will be close to that. Now go and try to figure out how to get the correct answer. Remember that cos x will simply fluctuate between -1 and 1 so keep that in mind when you calculate cos(100,000,000).
HallsofIvy
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Do you seriously want to ask only about 7- 2x? Yes, it should be obvious that will go to infinity as x goes to negative infinity. But the numerator, x, also goes to infinity and "infinity over infinity" is indeterminate.
With x going to infinity or negative infinity, a standard trick is to divide both numerator and denominator by the highest power of x. That way, every "x" becomes "1/x" which goes to 0 as x goes to inifnity. And "0" is much easier to work with!
For the fraction \(\displaystyle \frac{x}{7- 2x}\) divide both numerator and denominator by x so (as long as x is not 0 which is not a problem with x going to negative infinity) it becomes \(\displaystyle \frac{1}{\frac{7}{x}- 2}\). Now, what happens to \(\displaystyle \frac{7}{x}\) as x goes to negative infiity?
With x going to infinity or negative infinity, a standard trick is to divide both numerator and denominator by the highest power of x. That way, every "x" becomes "1/x" which goes to 0 as x goes to inifnity. And "0" is much easier to work with!
For the fraction \(\displaystyle \frac{x}{7- 2x}\) divide both numerator and denominator by x so (as long as x is not 0 which is not a problem with x going to negative infinity) it becomes \(\displaystyle \frac{1}{\frac{7}{x}- 2}\). Now, what happens to \(\displaystyle \frac{7}{x}\) as x goes to negative infiity?