What is the limit as x approaches 1 of (x^{2n}-2x^n+1)/(x^{2m)-2x^m+1} ?

The limit as x approaches 1 of (x^{2n}-2x^n+1)/(x^{2m)-2x^m+1} is (n^2)/(m^2)

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limit as x approaches 1 of (x^{2n}-2x^n+1)/(x^{2m)-2x^m+1}

x \to 1 lim (\frac{x ^{2 n} - 2 x ^{n} + 1}{x ^{2 m} - 2 x ^{m} + 1})

\frac{n ^{2}}{m ^{2}}

Solution steps

x \to 1 lim (\frac{x ^{2 n} - 2 x ^{n} + 1}{x ^{2 m} - 2 x ^{m} + 1})

Apply L'Hopital's Rule

: x \to 1 lim (\frac{2 n x ^{2 n - 1} - 2 n x ^{n - 1}}{2 m x ^{2 m - 1} - 2 m x ^{m - 1}})

= x \to 1 lim (\frac{2 n x ^{2 n - 1} - 2 n x ^{n - 1}}{2 m x ^{2 m - 1} - 2 m x ^{m - 1}})

Apply L'Hopital's Rule

: x \to 1 lim (\frac{2 n x ^{2 n - 2} ( 2 n - 1 ) - 2 n x ^{n - 2} ( n - 1 )}{2 m x ^{2 m - 2} ( 2 m - 1 ) - 2 m x ^{m - 2} ( m - 1 )})

= x \to 1 lim (\frac{2 n x ^{2 n - 2} ( 2 n - 1 ) - 2 n x ^{n - 2} ( n - 1 )}{2 m x ^{2 m - 2} ( 2 m - 1 ) - 2 m x ^{m - 2} ( m - 1 )})

Plug in the value

x = 1

= \frac{2 n \cdot 1 ^{2 n - 2} ( 2 n - 1 ) - 2 n \cdot 1 ^{n - 2} ( n - 1 )}{2 m \cdot 1 ^{2 m - 2} ( 2 m - 1 ) - 2 m \cdot 1 ^{m - 2} ( m - 1 )}

Simplify

\frac{2 n \cdot 1 ^{2 n - 2} ( 2 n - 1 ) - 2 n \cdot 1 ^{n - 2} ( n - 1 )}{2 m \cdot 1 ^{2 m - 2} ( 2 m - 1 ) - 2 m \cdot 1 ^{m - 2} ( m - 1 )} : \frac{n ^{2}}{m ^{2}}

= \frac{n ^{2}}{m ^{2}}

\int \frac{( 2 x ^{2} + 3 )}{x ( x - 1 ) ^{2}} d x

\int (cos (x) + 1)^{2} d x

\frac{d P}{d t} = Pt, P (1) = 2

\int 2 x + x^{2} d x

\frac{d}{d x} (\frac{2 x}{x})

What is the limit as x approaches 1 of (x^{2n}-2x^n+1)/(x^{2m)-2x^m+1} ?
The limit as x approaches 1 of (x^{2n}-2x^n+1)/(x^{2m)-2x^m+1} is (n^2)/(m^2)