Find the limit of $\frac{1}{\ln\left(x-6\right)}+\frac{-1}{x-7}$ as $x$ approaches $7$

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Final answer to the problem

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

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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

Learn how to solve limits by l'hôpital's rule problems step by step online.

$L.C.M.=\left(x-7\right)\ln\left(x-6\right)$

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Learn how to solve limits by l'hôpital's rule problems step by step online. Find the limit of 1/ln(x-6)+-1/(x-7) as x approaches 7. 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. 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. Combine and simplify all terms in the same fraction with common denominator \left(x-7\right)\ln\left(x-6\right). If we directly evaluate the limit \lim_{x\to7}\left(\frac{x-7-\ln\left(x-6\right)}{\left(x-7\right)\ln\left(x-6\right)}\right) as x tends to 7, we can see that it gives us an indeterminate form.

Final answer to the problem

$\frac{1}{2}$

Exact Numeric Answer

$0.5$

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Function Plot

Plotting: $\frac{1}{\ln\left(x-6\right)}+\frac{-1}{x-7}$

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7
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9
0
a
b
c
d
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

How to improve your answer:

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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