# 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 

How should I solve this problem?

• Choose an option
• Solve using L'H么pital's rule
• Solve without using l'H么pital
• Solve using limit properties
• Solve using direct substitution
• Solve the limit using factorization
• Solve the limit using rationalization
• Integrate by partial fractions
• Product of Binomials with Common Term
• FOIL Method
Can't find a method? Tell us so we can add it.
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 $\left(x-1\right)\ln\left(x\right)$

$\lim_{x\to1}\left(\frac{x\ln\left(x\right)-x+1}{\left(x-1\right)\ln\left(x\right)}\right)$
5

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

$\lim_{x\to1}\left(\frac{x\ln\left(x\right)-x+1}{x\ln\left(x\right)-\ln\left(x\right)}\right)$
6

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

$\frac{0}{0}$
7

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

$\lim_{x\to 1}\left(\frac{\frac{d}{dx}\left(x\ln\left(x\right)-x+1\right)}{\frac{d}{dx}\left(x\ln\left(x\right)-\ln\left(x\right)\right)}\right)$
8

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

$\lim_{x\to1}\left(\frac{\ln\left(x\right)}{\ln\left(x\right)+1+\frac{-1}{x}}\right)$
9

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

$\frac{0}{0}$
10

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

$\lim_{x\to 1}\left(\frac{\frac{d}{dx}\left(\ln\left(x\right)\right)}{\frac{d}{dx}\left(\ln\left(x\right)+1+\frac{-1}{x}\right)}\right)$
11

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

$\lim_{x\to1}\left(\frac{x}{1+x}\right)$
12

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

$\frac{1}{1+1}$
13

Add the values $1$ and $1$

$\frac{1}{2}$

##  Final answer to the problem

$\frac{1}{2}$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

SnapXam A2

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0
a
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f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

###  Main Topic: Limits by Direct Substitution

Find limits of functions at a specific point by directly plugging the value into the function.