# Compute the following limits using L'Hospital's rule if appropriate. Use INF to denote infinity and MINF to denote negative infinity   a)lim as x approaches 0 of (1-cos(3x))/(1-cos(5x)) b)lim as x approaches 1 of ((6^x)-(5^x)-(1))/((x^2)-1)

a) You need to substitute 0 for x in equation under limit such that:

`lim_(x-gt0)(1-cos(3x))/(1-cos(5x)) =(1-cos(0))/(1-cos(0)) = (1-1)/(1-1) = 0/0`

You should use l'Hospital's theorem such that:

`lim_(x-gt0) (1-cos(3x))/(1-cos(5x)) = lim_(x-gt0) ((1-cos(3x))')/((1-cos(5x))') `

`lim_(x-gt0) ((1-cos(3x))')/((1-cos(5x))') = lim_(x-gt0) (3sin(3x))/(5sin(5x)) = 0/0`

You need to use l'Hospital's theorem again such that:

`lim_(x-gt0) (3sin(3x))/(5sin(5x)) = lim_(x-gt0) ((3sin(3x))')/((5sin(5x)) ')`

`lim_(x-gt0) ((3sin(3x))')/((5sin(5x)) ') = lim_(x-gt0) (9cos(3x))/(25cos(5x))`

You need to substitute 0 for x in equation under limit such that:

`lim_(x-gt0) (9cos(3x))/(25cos(5x)) =(9cos(0))/(25cos(0))= 9/25`

Hence, evaluating the limit to the function yields `lim_(x-gt0) (1-cos(3x))/(1-cos(5x)) = 9/25.`

b) You need to substitute 1 for x in equation under limit such that:

`lim_(x-gt1)(6^x- 5^x - 1)/(x^2-1) = (6 - 5 - 1)/(1-1) = 0/0`

You should use l'Hospital's theorem such that:

`lim_(x-gt1)(6^x- 5^x - 1)/(x^2-1) =lim_(x-gt1)((6^x- 5^x - 1)')/((x^2-1)')`

`lim_(x-gt1)((6^x- 5^x - 1)')/((x^2-1)') = lim_(x-gt1)(6^xln 6- 5^xln 5)/(2x)`

You need to substitute 1 for x in equation under limit such that:

`lim_(x-gt1)(6^xln 6- 5^xln 5)/(2x) = (6ln 6- 5ln 5)/2`

Hence, evaluating the limit to the function yields  `lim_(x-gt1)(6^x- 5^x - 1)/(x^2-1) = ln sqrt (6^6/5^5).`

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