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Find the limit of $\frac{x}{x-1}+\frac{-1}{\ln\left(x\right)}$ as $x$ approaches $1$

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Final answer to the problem

$\frac{1}{2}$
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Step-by-step Solution

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  • Solve using L'Hôpital's rule
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  • Solve using limit properties
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1

The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors

$L.C.M.=\left(x-1\right)\ln\left(x\right)$
2

Obtained the least common multiple (LCM), we place it as the denominator of each fraction, and in the numerator of each fraction we add the factors that we need to complete

$\frac{x\ln\left(x\right)}{\left(x-1\right)\ln\left(x\right)}+\frac{-\left(x-1\right)}{\left(x-1\right)\ln\left(x\right)}$
3

Simplify the numerators

$\frac{x\ln\left(x\right)}{\left(x-1\right)\ln\left(x\right)}+\frac{-x+1}{\left(x-1\right)\ln\left(x\right)}$
4

Combine and simplify all terms in the same fraction with common denominator $\left(x-1\right)\ln\left(x\right)$

$\lim_{x\to1}\left(\frac{x\ln\left(x\right)-x+1}{\left(x-1\right)\ln\left(x\right)}\right)$

Plug in the value $1$ into the limit

$\lim_{x\to1}\left(\frac{1\ln\left(1\right)- 1+1}{\left(1-1\right)\ln\left(1\right)}\right)$

Multiply $-1$ times $1$

$\lim_{x\to1}\left(\frac{1\ln\left(1\right)-1+1}{\left(1-1\right)\ln\left(1\right)}\right)$

Subtract the values $1$ and $-1$

$\lim_{x\to1}\left(\frac{1\ln\left(1\right)}{\left(1-1\right)\ln\left(1\right)}\right)$

Calculating the natural logarithm of $1$

$\lim_{x\to1}\left(\frac{1\cdot 0}{\left(1-1\right)\ln\left(1\right)}\right)$

Multiply $1$ times $0$

$\lim_{x\to1}\left(\frac{0}{\left(1-1\right)\ln\left(1\right)}\right)$

Subtract the values $1$ and $-1$

$\lim_{x\to1}\left(\frac{0}{0\ln\left(1\right)}\right)$

Calculating the natural logarithm of $1$

$\lim_{x\to1}\left(\frac{0}{0\cdot 0}\right)$

Multiply $0$ times $0$

$\lim_{x\to1}\left(\frac{0}{0}\right)$
5

If we directly evaluate the limit $\lim_{x\to1}\left(\frac{x\ln\left(x\right)-x+1}{\left(x-1\right)\ln\left(x\right)}\right)$ as $x$ tends to $1$, we can see that it gives us an indeterminate form

$\frac{0}{0}$
6

We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

$\lim_{x\to 1}\left(\frac{\frac{d}{dx}\left(x\ln\left(x\right)-x+1\right)}{\frac{d}{dx}\left(\left(x-1\right)\ln\left(x\right)\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(x\ln\left(x\right)-x+1\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(x\ln\left(x\right)\right)+\frac{d}{dx}\left(-x\right)$

The derivative of the linear function times a constant, is equal to the constant

$\frac{d}{dx}\left(x\ln\left(x\right)\right)-\frac{d}{dx}\left(x\right)$

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=x$ and $g=\ln\left(x\right)$

$\frac{d}{dx}\left(x\right)\ln\left(x\right)+x\frac{d}{dx}\left(\ln\left(x\right)\right)-\frac{d}{dx}\left(x\right)$

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{d}{dx}\left(x\right)\ln\left(x\right)+x\frac{1}{x}-\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$\ln\left(x\right)+x\frac{1}{x}-1$

Multiplying the fraction by $x$

$\ln\left(x\right)+\frac{x}{x}-1$

Simplify the fraction $\frac{x}{x}$ by $x$

$\ln\left(x\right)+1-1$

Add the values $1$ and $-1$

$\ln\left(x\right)+0$

$x+0=x$, where $x$ is any expression

$\ln\left(x\right)$

Find the derivative of the denominator

$\frac{d}{dx}\left(\left(x-1\right)\ln\left(x\right)\right)$

Multiply the single term $\ln\left(x\right)$ by each term of the polynomial $\left(x-1\right)$

$\frac{d}{dx}\left(x\ln\left(x\right)-\ln\left(x\right)\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(x\ln\left(x\right)\right)+\frac{d}{dx}\left(-\ln\left(x\right)\right)$

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$\frac{d}{dx}\left(x\ln\left(x\right)\right)-\frac{d}{dx}\left(\ln\left(x\right)\right)$

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=x$ and $g=\ln\left(x\right)$

$\frac{d}{dx}\left(x\right)\ln\left(x\right)+x\frac{d}{dx}\left(\ln\left(x\right)\right)-\frac{d}{dx}\left(\ln\left(x\right)\right)$

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{d}{dx}\left(x\right)\ln\left(x\right)+x\frac{1}{x}-\frac{d}{dx}\left(\ln\left(x\right)\right)$

The derivative of the linear function is equal to $1$

$\ln\left(x\right)+x\frac{1}{x}-\frac{d}{dx}\left(\ln\left(x\right)\right)$

Multiplying the fraction by $x$

$\ln\left(x\right)+\frac{x}{x}-\frac{d}{dx}\left(\ln\left(x\right)\right)$

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\ln\left(x\right)+\frac{x}{x}+\frac{-1}{x}$

Simplify the fraction $\frac{x}{x}$ by $x$

$\ln\left(x\right)+1+\frac{-1}{x}$
7

After deriving both the numerator and denominator, and simplifying, the limit results in

$\lim_{x\to1}\left(\frac{\ln\left(x\right)}{\ln\left(x\right)+1+\frac{-1}{x}}\right)$

Plug in the value $1$ into the limit

$\lim_{x\to1}\left(\frac{\ln\left(1\right)}{\ln\left(1\right)+1-\frac{1}{1}}\right)$

Calculating the natural logarithm of $1$

$\lim_{x\to1}\left(\frac{0}{\ln\left(1\right)+1-\frac{1}{1}}\right)$

Divide $-1$ by $1$

$\lim_{x\to1}\left(\frac{0}{\ln\left(1\right)+1-1}\right)$

Subtract the values $1$ and $-1$

$\lim_{x\to1}\left(\frac{0}{\ln\left(1\right)}\right)$

Calculating the natural logarithm of $1$

$\lim_{x\to1}\left(\frac{0}{0}\right)$
8

If we directly evaluate the limit $\lim_{x\to1}\left(\frac{\ln\left(x\right)}{\ln\left(x\right)+1+\frac{-1}{x}}\right)$ as $x$ tends to $1$, we can see that it gives us an indeterminate form

$\frac{0}{0}$
9

We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

$\lim_{x\to 1}\left(\frac{\frac{d}{dx}\left(\ln\left(x\right)\right)}{\frac{d}{dx}\left(\ln\left(x\right)+1+\frac{-1}{x}\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(\ln\left(x\right)\right)$

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{x}$

Find the derivative of the denominator

$\frac{d}{dx}\left(\ln\left(x\right)+1+\frac{-1}{x}\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(\ln\left(x\right)\right)+\frac{d}{dx}\left(\frac{-1}{x}\right)$

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{x}+\frac{d}{dx}\left(\frac{-1}{x}\right)$

Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{1}{x}+\frac{\frac{d}{dx}\left(-1\right)x- -\frac{d}{dx}\left(x\right)}{x^2}$

Multiply $-1$ times $-1$

$\frac{1}{x}+\frac{\frac{d}{dx}\left(-1\right)x+\frac{d}{dx}\left(x\right)}{x^2}$

The derivative of the constant function ($-1$) is equal to zero

$\frac{1}{x}+\frac{0+\frac{d}{dx}\left(x\right)}{x^2}$

The derivative of the linear function is equal to $1$

$\frac{1}{x}+\frac{0+1}{x^2}$

$x+0=x$, where $x$ is any expression

$\frac{1}{x}+\frac{1}{x^2}$

Combine $\frac{1}{x}+\frac{1}{x^2}$ in a single fraction

$\lim_{x\to1}\left(\frac{\frac{1}{x}}{\frac{1+\frac{x^2}{x}}{x^2}}\right)$

Divide fractions $\frac{\frac{1}{x}}{\frac{1+\frac{x^2}{x}}{x^2}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$\lim_{x\to1}\left(\frac{\frac{x^2}{x}}{1+\frac{x^2}{x}}\right)$

Simplify the fraction $\frac{x^2}{x}$ by $x$

$\lim_{x\to1}\left(\frac{x}{1+\frac{x^2}{x}}\right)$

Simplify the fraction $\frac{x^2}{x}$ by $x$

$\lim_{x\to1}\left(\frac{x}{1+x}\right)$
10

After deriving both the numerator and denominator, and simplifying, the limit results in

$\lim_{x\to1}\left(\frac{x}{1+x}\right)$

Evaluate the limit $\lim_{x\to1}\left(\frac{x}{1+x}\right)$ by replacing all occurrences of $x$ by $1$

$\frac{1}{1+1}$

Add the values $1$ and $1$

$\frac{1}{2}$
11

Evaluate the limit $\lim_{x\to1}\left(\frac{x}{1+x}\right)$ by replacing all occurrences of $x$ by $1$

$\frac{1}{2}$

Final answer to the problem

$\frac{1}{2}$

Exact Numeric Answer

$0.5$

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Function Plot

Plotting: $\frac{x}{x-1}+\frac{-1}{\ln\left(x\right)}$

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log◻
lim
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θ
=
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<
>=
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sin
cos
tan
cot
sec
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asin
acos
atan
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asec
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sinh
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tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Limits by L'Hôpital's rule

In mathematics, and more specifically in calculus, L'Hôpital's rule uses derivatives to help evaluate limits involving indeterminate forms.

Used Formulas

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