Nickolas Taylor

Answered

2022-07-09

Rewrite fraction to calculate limit

I am practising finding limits. However, I can't seem to figure out this one.

$f(x)=\frac{{x}^{3}+4x-5}{{x}^{2}-1}\text{as}x\text{goes to}1$

I understand I have to rewrite the fraction somehow for the denominator not to equal 0, but I don't know where to start.

I am practising finding limits. However, I can't seem to figure out this one.

$f(x)=\frac{{x}^{3}+4x-5}{{x}^{2}-1}\text{as}x\text{goes to}1$

I understand I have to rewrite the fraction somehow for the denominator not to equal 0, but I don't know where to start.

Answer & Explanation

Brendan Bush

Expert

2022-07-10Added 14 answers

Using the Euclidean division of ${x}^{3}+4x-5$ by $x-1$ we get

$f(x)=\frac{{x}^{3}+4x-5}{{x}^{2}-1}=\frac{(x-1)({x}^{2}+x+5)}{(x-1)(x+1)}$

$f(x)=\frac{{x}^{3}+4x-5}{{x}^{2}-1}=\frac{(x-1)({x}^{2}+x+5)}{(x-1)(x+1)}$

Cooper Doyle

Expert

2022-07-11Added 2 answers

One idea is to use polynomial long division.

The idea is to note that you have a cubic divided by a quadratic, so the degree of the numerator is greater by 1. Consequently, we can conclude that

$f(x)=ax+b+\frac{cx+d}{{x}^{2}-1}$

for some constants $a,b,c,d,$ where the linear numerator $cx+d$ is to allow for the fact that there may be a remainder term, which is necessarily of lower degree than the denominator.

Multiplying both sides of this equation by ${x}^{2}-1$ -which is non-0 for x sufficiently close (but not equal) to 1--we obtain

${x}^{3}+4x-5=(ax+b)({x}^{2}-1)+cx+d.$

Expand the product on the right-hand side to give yourself a system of equations. Solve for $a,b,c,d.$

Once you've found these, the rest should fall right out of your usual limit manipulations.

The idea is to note that you have a cubic divided by a quadratic, so the degree of the numerator is greater by 1. Consequently, we can conclude that

$f(x)=ax+b+\frac{cx+d}{{x}^{2}-1}$

for some constants $a,b,c,d,$ where the linear numerator $cx+d$ is to allow for the fact that there may be a remainder term, which is necessarily of lower degree than the denominator.

Multiplying both sides of this equation by ${x}^{2}-1$ -which is non-0 for x sufficiently close (but not equal) to 1--we obtain

${x}^{3}+4x-5=(ax+b)({x}^{2}-1)+cx+d.$

Expand the product on the right-hand side to give yourself a system of equations. Solve for $a,b,c,d.$

Once you've found these, the rest should fall right out of your usual limit manipulations.

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