Limit Problem - # 4

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Jason76

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\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

Next move:confused:

Could it be?

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\) :confused:
 
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Jason76 said:
\(\displaystyle \displaystyle{\lim_{x \rightarrow 1}\, \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}}\)

Could it be?

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\) :confused:
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What did you get when you combined the fractions into one term? Did you apply l'Hospital's Rule? ;)
 
Jason76 said:
\(\displaystyle \displaystyle \lim _{x \to 1}\left(\dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\right)\)

........\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\) ..X what happened to the "x" from the 1st numerator?

........\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\) :confused:
Click to expand...
After combining the two fractions over the LCD, what do you have?
 
\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

Next move:confused:

Could it be?

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5x \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5x \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\)

Not sure what the subtraction on the top will lead to. Any hints?
 
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\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{\ln x^{2} - \ln x}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} 5 \ln x - 5x - 5}{\dfrac{d}{dx} 2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{2}{x} - \dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\)
 
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Jason76 said:
\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{\ln x^{2} - \ln x}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} 5 \ln x - 5x - 5}{\dfrac{d}{dx} 2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{2}{x} - \dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\) :confused: Answer?
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That is not yet the answer. If you were writing correct statements, you would see that you have not yet taken the limit as x-->1

\(\displaystyle \displaystyle \lim_{x\to 1}\dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\)
 
\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{\ln x^{2} - \ln x}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} 5 \ln x - 5x - 5}{\dfrac{d}{dx} 2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{2}{x} - \dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{(1)} - 5}{\dfrac{1}{(1)}} = \dfrac{0}{0}\) Indeterminate

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} \dfrac{5}{x} - 5}{\dfrac{d}{dx}\dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{(x)(0) - 5(1)}{x^{2}}}{\dfrac{(x)(0) - 1(1)}{x^{2}}}\)

\(\displaystyle \dfrac{\dfrac{-5}{x^{2}}}{\dfrac{-1}{x^{2}}}\)

\(\displaystyle \dfrac{\dfrac{-5}{(1)^{2}}}{\dfrac{-1}{(1)^{2}}} = \dfrac{0}{0}\) Indeterminate
 
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Finally, this looks right:

\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{\ln x^{2} - \ln x}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} 5 \ln x - 5x - 5}{\dfrac{d}{dx} 2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{2}{x} - \dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{(1)} - 5}{\dfrac{1}{(1)}} = \dfrac{0}{1} = 0\) Answer?? Computer still says no.
 
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Jason76 said:
Finally, this looks right:

\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{\ln x^{2} - \ln x}\)............................................Where did this come from??!!

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} 5 \ln x - 5x - 5}{\dfrac{d}{dx} 2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{2}{x} - \dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{(1)} - 5}{\dfrac{1}{(1)}} = \dfrac{0}{1} = 0\) Answer?? Computer still says no.
Click to expand...

Horrible algebra!!!
 
Jason76 said:
Finally, this looks right:

\(\displaystyle \lim x \rightarrow 1\)

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}\)

\(\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}\) Indeterminate

\(\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}\)

\(\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}\)
Click to expand...
-(5x- 5)= -5x+ 5

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{\ln x^{2} - \ln x}\)

\(\displaystyle \dfrac{5 \ln x - 5x - 5}{2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} 5 \ln x - 5x - 5}{\dfrac{d}{dx} 2 \ln x - \ln x}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{2}{x} - \dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}\)

\(\displaystyle \dfrac{\dfrac{5}{(1)} - 5}{\dfrac{1}{(1)}} = \dfrac{0}{1} = 0\) Answer?? Computer still says no.
Click to expand...
 
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