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!
Find the limit: $\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$
2
$\begingroup$
calculus
limits
limits-without-lhopital
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0you can not use L'Hospital? – 2017-01-28
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0@Dr.SonnhardGraubner No, I can't. – 2017-01-28
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2Hint: $\lim_{x \to 0} (a^x - 1) / x = \ln a\,$. – 2017-01-28
3 Answers
3
$$\lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} = \lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} \cdot \frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{6^x-1-(5^x-1)}\cdot\frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{x}\cdot\lim_{x\to0}\frac{x}{6^x-1-(5^x-1)} = \bigg(\lim_{x\to0}\frac{8^x-1}{x}-\frac{7^x-1}{x}\bigg)\cdot\lim_{x\to0} \frac{1}{\frac{6^x-1}{x}-\frac{5^x-1}{x}} = (\ln8-\ln7)\cdot\frac{1}{\ln6-\ln5} = \frac{\ln\frac{8}{7}}{\ln\frac{6}{5}}$$
8
Since $a^x =e^{x \ln a} \approx 1+ x\ln a $ as $x \to 0$. the limit is
$\dfrac{\ln(8/7)}{\ln (6/5)}$
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2Concise and clear. (+1) – 2017-01-28
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1It had to be. I entered it from my phone. – 2017-01-29
3
By using hint of @dxix we get $$\lim _{ x\to 0 } \frac { 8^{ x }-7^{ x } }{ 6^{ x }-5^{ x } } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\left[ { \left( \frac { 8 }{ 7 } \right) }^{ x }-1 \right] }{ { 5 }^{ x }\left[ { \left( \frac { 6 }{ 5 } \right) }^{ x }-1 \right] } =\lim _{ x\to 0 } \frac { { 7 }^{ x }\frac { \left[ { \left( \frac { 8 }{ 7 } \right) }^{ x }-1 \right] }{ x } }{ { 5 }^{ x }\frac { \left[ { \left( \frac { 6 }{ 5 } \right) }^{ x }-1 \right] }{ x } } =\\ =\frac { \ln { \left( 8/7 \right) } }{ \ln { \left( 6/5 \right) } } $$