Find the limit: $\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$ Find the limit:
$$\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$$
I really have no idea what to do here, since I obviously can't plug in $x=0$ nor divide by highest power...Any help is appreciated!
 A: Since
$a^x =e^{x \ln a}
\approx 1+ x\ln a
$ as
$x \to 0$.
the limit is
$\dfrac{\ln(8/7)}{\ln (6/5)}$
A: By using hint of @dxix we get 
$$\lim _{ x\to 0 } \frac { 8^{ x }-7^{ x } }{ 6^{ x }-5^{ x } } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\left[ { \left( \frac { 8 }{ 7 }  \right)  }^{ x }-1 \right]  }{ { 5 }^{ x }\left[ { \left( \frac { 6 }{ 5 }  \right)  }^{ x }-1 \right]  } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\frac { \left[ { \left( \frac { 8 }{ 7 }  \right)  }^{ x }-1 \right]  }{ x }  }{ { 5 }^{ x }\frac { \left[ { \left( \frac { 6 }{ 5 }  \right)  }^{ x }-1 \right]  }{ x }  } =\\ =\frac { \ln { \left( 8/7 \right)  }  }{ \ln { \left( 6/5 \right)  }  } $$
A: $$\lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} = \lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} \cdot \frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{6^x-1-(5^x-1)}\cdot\frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{x}\cdot\lim_{x\to0}\frac{x}{6^x-1-(5^x-1)} = \bigg(\lim_{x\to0}\frac{8^x-1}{x}-\frac{7^x-1}{x}\bigg)\cdot\lim_{x\to0} \frac{1}{\frac{6^x-1}{x}-\frac{5^x-1}{x}} = (\ln8-\ln7)\cdot\frac{1}{\ln6-\ln5} = \frac{\ln\frac{8}{7}}{\ln\frac{6}{5}}$$
