find the limit. use ihospitals rule where appropriate. if there is a more elementary method, consider using…

find the limit. use ihospitals rule where appropriate. if there is a more elementary method, consider using it. \n\n\\( \\lim _ { x \\rightarrow 0 } \\frac { x 2 ^ { x } } { 2 ^ { x } - 1 } \\)\n\n14. -/1 points\n\nfind the limit. use ihospitals rule where appropriate. if there is a more elementary method, consider using it. \n\n\\( \\lim _ { x \\rightarrow 0 ^ { + } } \\frac { \\arctan ( 8 x ) } { \\ln ( x ) } \\)

find the limit. use ihospitals rule where appropriate. if there is a more elementary method, consider using it. \n\n\\( \\lim _ { x \\rightarrow 0 } \\frac { x 2 ^ { x } } { 2 ^ { x } - 1 } \\)\n\n14. -/1 points\n\nfind the limit. use ihospitals rule where appropriate. if there is a more elementary method, consider using it. \n\n\\( \\lim _ { x \\rightarrow 0 ^ { + } } \\frac { \\arctan ( 8 x ) } { \\ln ( x ) } \\)

Answer

Explanation:

Step1: Check the form of the limit

When (x\rightarrow0^{+}), (\arctan(8x)\rightarrow0) and (\ln(x)\rightarrow-\infty). The limit (\lim_{x\rightarrow0^{+}}\frac{\arctan(8x)}{\ln(x)}) is in the form (\frac{0}{-\infty}).

Step2: Apply the property of limits

We know that if (\lim_{x\rightarrow a}f(x) = 0) and (\lim_{x\rightarrow a}|g(x)|=\infty), then (\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = 0). Here (f(x)=\arctan(8x)), (g(x)=\ln(x)) and (a = 0^{+}).

Answer:

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