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May 4th, 2018, 10:18 PM  #1 
Newbie Joined: May 2018 From: China Posts: 3 Thanks: 0  please help with these questions on Limits
Need help urgently

May 4th, 2018, 10:30 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,299 Thanks: 1218 
there is nothing particularly difficult about any of these problems. If you show what work you've done on them we can help you with any difficulties you have. 
May 4th, 2018, 11:35 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
$$\lim_{x \to \infty} \frac{5x^3+11}{4x+3}$$ As $x$ grows without bound, the numerator $(5x^3+11)$ and the denominator $(4x+3)$ both grow without bound (thinking in terms of magnitude, the sign isn't important here). This means that the fraction ultimately is $\frac\infty\infty$ which we can't calculate because it's an indeterminate form. So we need to find a way of modifying both the numerator and the denominator so that
The third condition is necessary because $\frac00$ is just another indeterminate form, so we'd have made no progress. The first condition is achieved by multiplying (or dividing) the numerator by the same factor as the denominator (because $\frac{ac}{bc} = \frac{a}{b}$). All we need to do is to determine the relevant factor. So we have to find a single factor by which we can multiply both numerator and denominator so that both are finite for large values of $x$. If we choose $\frac1x$, we see that the denominator becomes $$\frac1x(4x+3) = \frac{4x}{x} + \frac3x = 4 + \frac3x$$ and this has a finite limit (4) as $x$ grows without bound. But the numerator, multiplied by the same factor becomes $$\frac1x(5x^3+11) = \frac5x^2 + \frac{11}{x}$$ which grows without bound with $x$. So the factor $\frac1x$ won't work. If we choose the factor $\frac1{x^3}$, the denominator becomes $$\frac1{x^3}(4x+3) = \frac4{x^2} + \frac3{x^3}$$ which has a finite limit of zero. Meanwhile the numerator becomes $$\frac1{x^3}(5x^3+11) = \frac5 + \frac{11}{x^3}$$ which has a finite limit of 5. So this factor works fine. You will see that the factor $\frac1{x^4}$ sends both the numerator and the denominator to zero in the limit. That violates the third condition, so that factor doesn't work. In general, for rational functions of polynomials, the factor we want is the reciprocal of the greatest power of $x$ that appears in either the numerator or the denominator (or both). In this case we use the factor $\frac1{x^3}$ and get \begin{align*} \lim_{x \to \infty} \frac{5x^3 + 11}{4x+3} &= \lim_{x \to \infty} \frac{\frac{1}{x^3}(5x^3 + 11)}{\frac{1}{x^3}(4x+3)} \\ &= \lim_{x \to \infty} \frac{5 + \frac{11}{x^3}}{\frac{4}{x^2}+\frac{3}{x^3}} \\ &\to \frac{5 + 0}{0+0} \end{align*} Since the numerator is positive and the denominator is zero, the limit is thus $+\infty$. I'll leave the others for you. 
May 5th, 2018, 05:20 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
Or you can use the general rule: the limit, as x goes to infinity, of one polynomial divided by another polynomial is a) 0 if the polynomial in the denominator has higher degree than the polynomial in the numerator. b) the ratio of the leading coefficients if the degrees are the same. c) plus or minus infinity if the polynomial in the numerator has higher degree, the sign being the same as the ratio of leading coefficients. 
May 6th, 2018, 07:21 AM  #5 
Newbie Joined: May 2018 From: China Posts: 3 Thanks: 0 
thanks, that helped...managed to get on with the other questions 

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