Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 1} (\frac{x}{x-1}-\frac{1}{\ln x})
Final Answer
Solution
First, we try to plug in x = 1 and get
(\frac{1}{1-1}-\frac{1}{\ln 1})=
=(\frac{1}{0}-\frac{1}{0})=
=(\infty-\infty)
Note: The zero in the denominator is not absolute zero, but a number that is tending to zero.
We got the phrase \infty-\infty (=infinity minus infinity). This is an indeterminate form, therefore we have to get out of this situation.
\lim _ { x \rightarrow 1} (\frac{x}{x-1}-\frac{1}{\ln x})=
In order to use Lopital Rule, we will break down the polynomials into factors and calculate the least common denominator
=\lim _ { x \rightarrow 1} \frac{x\ln x-(x-1)}{(x-1)\ln x}=
=\lim _ { x \rightarrow 1} \frac{x\ln x-x+1}{x\ln x-\ln x}=
We plug in one again and get
= \frac{1\cdot\ln 1-1+1}{1\cdot \ln 1-\ln 1}=
= \frac{0}{0}
We got the phrase \frac{"0"}{"0"}(=tending to zero divides tending to zero). This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get
=\lim _ { x \rightarrow 1} \frac{\ln x+1-1}{\ln x+1-\frac{1}{x}}=
We simplify the expression and get
=\lim _ { x \rightarrow 1} \frac{\ln x}{\ln x+1-\frac{1}{x}}=
We plug in one again and get
=\frac{\ln 1}{\ln 1+1-\frac{1}{1}}=
=\frac{0}{0}
We got the phrase \frac{0}{0} again, therefore we use Lopital Rule again – we derive the numerator and denominator separately and we will get
=\lim _ { x \rightarrow 1} \frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^2}}=
We simplify the expression and get
=\lim _ { x \rightarrow 1} \frac{x}{x+1}=
We plug in one again and this time we get
=\frac{1}{1+1}=
=\frac{1}{2}
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