Find the limit $lim_{x to 0} (frac{1}{ln (x+sqrt{1+x^2)} }- frac{1}{ln( x+1)} ) =?$ [duplicate]
This question already has an answer here:
Proving $limlimits_{xto0}left(frac{1}{log(x+sqrt{1+x^2})}-frac{1}{log(1+x)}right) =-frac12$
2 answers
Find the limit :
$$lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2}) }- frac{1}{ln( x+1)} right) =?$$
My Try :
$$lim_{x to 0} left(frac{ln(x+1)-ln (x+sqrt{1+x^2})}{ln (x+sqrt{1+x^2})ln(x+1) }right)=?$$
$$lim_{x to 0} dfrac{lnleft(dfrac{x+1}{x+sqrt{1+x^2}}right)}{ln (x+sqrt{1+x^2})ln(x+1) }=?$$
Now what ?
calculus limits
edited 2 days ago
Siong Thye Goh
98.9k1464116
asked Dec 26 at 0:34
Almot1960
2,487723
marked as duplicate by Paramanand Singh calculus
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Proving $limlimits_{xto0}left(frac{1}{log(x+sqrt{1+x^2})}-frac{1}{log(1+x)}right) =-frac12$
2 answers
Find the limit :
$$lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2}) }- frac{1}{ln( x+1)} right) =?$$
My Try :
$$lim_{x to 0} left(frac{ln(x+1)-ln (x+sqrt{1+x^2})}{ln (x+sqrt{1+x^2})ln(x+1) }right)=?$$
$$lim_{x to 0} dfrac{lnleft(dfrac{x+1}{x+sqrt{1+x^2}}right)}{ln (x+sqrt{1+x^2})ln(x+1) }=?$$
Now what ?
calculus limits
edited 2 days ago
Siong Thye Goh
98.9k1464116
asked Dec 26 at 0:34
Almot1960
2,487723
marked as duplicate by Paramanand Singh calculus
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2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Proving $limlimits_{xto0}left(frac{1}{log(x+sqrt{1+x^2})}-frac{1}{log(1+x)}right) =-frac12$
2 answers
Find the limit :
$$lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2}) }- frac{1}{ln( x+1)} right) =?$$
My Try :
$$lim_{x to 0} left(frac{ln(x+1)-ln (x+sqrt{1+x^2})}{ln (x+sqrt{1+x^2})ln(x+1) }right)=?$$
$$lim_{x to 0} dfrac{lnleft(dfrac{x+1}{x+sqrt{1+x^2}}right)}{ln (x+sqrt{1+x^2})ln(x+1) }=?$$
Now what ?
calculus limits
edited 2 days ago
Siong Thye Goh
98.9k1464116
asked Dec 26 at 0:34
Almot1960
2,487723
This question already has an answer here:
Proving $limlimits_{xto0}left(frac{1}{log(x+sqrt{1+x^2})}-frac{1}{log(1+x)}right) =-frac12$
2 answers
Find the limit :
$$lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2}) }- frac{1}{ln( x+1)} right) =?$$
My Try :
$$lim_{x to 0} left(frac{ln(x+1)-ln (x+sqrt{1+x^2})}{ln (x+sqrt{1+x^2})ln(x+1) }right)=?$$
$$lim_{x to 0} dfrac{lnleft(dfrac{x+1}{x+sqrt{1+x^2}}right)}{ln (x+sqrt{1+x^2})ln(x+1) }=?$$
Now what ?
This question already has an answer here:
Proving $limlimits_{xto0}left(frac{1}{log(x+sqrt{1+x^2})}-frac{1}{log(1+x)}right) =-frac12$
2 answers
calculus limits
calculus limits
edited 2 days ago
Siong Thye Goh
98.9k1464116
asked Dec 26 at 0:34
Almot1960
2,487723
edited 2 days ago
Siong Thye Goh
98.9k1464116
asked Dec 26 at 0:34
Almot1960
2,487723
edited 2 days ago
Siong Thye Goh
98.9k1464116
edited 2 days ago
Siong Thye Goh
98.9k1464116
edited 2 days ago
Siong Thye Goh
98.9k1464116
98.9k1464116
asked Dec 26 at 0:34
Almot1960
2,487723
asked Dec 26 at 0:34
Almot1960
2,487723
asked Dec 26 at 0:34
Almot1960
2,487723
2,487723
marked as duplicate by Paramanand Singh calculus
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2 Answers
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From Taylor expansion,$$ln (x+sqrt{1+x^2})approx ln (1+x+frac12x^2)approx x+frac12x^2-frac12x^2 approx x$$
$$ln (1+x) approx x-frac{x^2}2$$
begin{align}
lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2)} }- frac{1}{ln( x+1)} right)&=lim_{x to 0} left(frac1{x} - frac1{x-frac{x^2}2} right)\
&=lim_{x to 0} left(frac{-frac{x^2}2}{x(x-frac{x^2}2)} right)\
&=lim_{x to 0} left( frac{-frac12}{1-frac{x}2}right)\
&= -frac12
end{align}
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
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Hint: See zero in the limit? See log, trig, exponents or roots? Taylor is your best friend. Apply it to the original expression, of course.
answered Dec 26 at 0:50
Makina
1,163115
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2 Answers
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2 Answers
2
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active
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From Taylor expansion,$$ln (x+sqrt{1+x^2})approx ln (1+x+frac12x^2)approx x+frac12x^2-frac12x^2 approx x$$
$$ln (1+x) approx x-frac{x^2}2$$
begin{align}
lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2)} }- frac{1}{ln( x+1)} right)&=lim_{x to 0} left(frac1{x} - frac1{x-frac{x^2}2} right)\
&=lim_{x to 0} left(frac{-frac{x^2}2}{x(x-frac{x^2}2)} right)\
&=lim_{x to 0} left( frac{-frac12}{1-frac{x}2}right)\
&= -frac12
end{align}
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
add a comment |
From Taylor expansion,$$ln (x+sqrt{1+x^2})approx ln (1+x+frac12x^2)approx x+frac12x^2-frac12x^2 approx x$$
$$ln (1+x) approx x-frac{x^2}2$$
begin{align}
lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2)} }- frac{1}{ln( x+1)} right)&=lim_{x to 0} left(frac1{x} - frac1{x-frac{x^2}2} right)\
&=lim_{x to 0} left(frac{-frac{x^2}2}{x(x-frac{x^2}2)} right)\
&=lim_{x to 0} left( frac{-frac12}{1-frac{x}2}right)\
&= -frac12
end{align}
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
add a comment |
From Taylor expansion,$$ln (x+sqrt{1+x^2})approx ln (1+x+frac12x^2)approx x+frac12x^2-frac12x^2 approx x$$
$$ln (1+x) approx x-frac{x^2}2$$
begin{align}
lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2)} }- frac{1}{ln( x+1)} right)&=lim_{x to 0} left(frac1{x} - frac1{x-frac{x^2}2} right)\
&=lim_{x to 0} left(frac{-frac{x^2}2}{x(x-frac{x^2}2)} right)\
&=lim_{x to 0} left( frac{-frac12}{1-frac{x}2}right)\
&= -frac12
end{align}
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
From Taylor expansion,$$ln (x+sqrt{1+x^2})approx ln (1+x+frac12x^2)approx x+frac12x^2-frac12x^2 approx x$$
$$ln (1+x) approx x-frac{x^2}2$$
begin{align}
lim_{x to 0} left(frac{1}{ln (x+sqrt{1+x^2)} }- frac{1}{ln( x+1)} right)&=lim_{x to 0} left(frac1{x} - frac1{x-frac{x^2}2} right)\
&=lim_{x to 0} left(frac{-frac{x^2}2}{x(x-frac{x^2}2)} right)\
&=lim_{x to 0} left( frac{-frac12}{1-frac{x}2}right)\
&= -frac12
end{align}
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
answered Dec 26 at 1:11
Siong Thye Goh
98.9k1464116
98.9k1464116
add a comment |
add a comment |
Hint: See zero in the limit? See log, trig, exponents or roots? Taylor is your best friend. Apply it to the original expression, of course.
answered Dec 26 at 0:50
Makina
1,163115
add a comment |
Hint: See zero in the limit? See log, trig, exponents or roots? Taylor is your best friend. Apply it to the original expression, of course.
answered Dec 26 at 0:50
Makina
1,163115
add a comment |
Hint: See zero in the limit? See log, trig, exponents or roots? Taylor is your best friend. Apply it to the original expression, of course.
answered Dec 26 at 0:50
Makina
1,163115
Hint: See zero in the limit? See log, trig, exponents or roots? Taylor is your best friend. Apply it to the original expression, of course.
answered Dec 26 at 0:50
Makina
1,163115
answered Dec 26 at 0:50
Makina
1,163115
answered Dec 26 at 0:50
Makina
1,163115
answered Dec 26 at 0:50
Makina
1,163115
1,163115
add a comment |
add a comment |
