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How do you solve #(m^2 + 2m + 1)/( m^3 + 3m^2 + 3m +1)(m^2/(m - (3m)/3)) #?

Answer 1

Eva Adamson

The function doesn't exist.

#color(white)=(m^2+2m+1)/(m^3+3m^2+3m 1)(m^2/(m-(3m)/3))#

#=(m^2+2m+1)/(m^3+3m^2+3m 1)(m^2/(m-(color(red)cancelcolor(black)3m)/color(red)cancelcolor(black)3))#

#=(m^2+2m+1)/(m^3+3m^2+3m 1)(m^2/(color(red)cancelcolor(black)(m-m)))#

#=(m^2+2m+1)/(m^3+3m^2+3m 1)(m^2/0)#

Since there is division by zero, the function cannot exist.

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Answer 2

Eva Abramson

To solve the expression (m^2 + 2m + 1)/(m^3 + 3m^2 + 3m + 1)(m^2/(m - (3m)/3)), we can simplify it step by step.

First, let's simplify the expression in the denominator: (m^3 + 3m^2 + 3m + 1)(m^2/(m - (3m)/3)). Simplifying the denominator, we have (m^3 + 3m^2 + 3m + 1)(m^2/(m - m)). Simplifying further, we get (m^3 + 3m^2 + 3m + 1)(m^2/0). Since division by zero is undefined, the expression is undefined.

Therefore, the given expression cannot be solved as it leads to an undefined result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.