How to make this limit question a indeterminate form? (L-Hopital)
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$$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$
This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.
How to solve this question?
limits
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:35
Amogh Joshi
183
add a comment |
up vote
3
down vote
favorite
$$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$
This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.
How to solve this question?
limits
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:35
Amogh Joshi
183
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$
This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.
How to solve this question?
limits
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:35
Amogh Joshi
183
$$lim_{xto 1^+}[ln(x^7 -1) - ln(x^5 -1)]$$
This is a question from L-Hospital rule question set. My approach was to apply log property in this question and solve it, but $ln(frac{0}{0})$ might not be right way to convert it into indeterminate form.
How to solve this question?
limits
limits
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:35
Amogh Joshi
183
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
asked Nov 14 at 6:35
Amogh Joshi
183
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
edited Nov 14 at 6:52
Ng Chung Tak
13.6k31234
13.6k31234
asked Nov 14 at 6:35
Amogh Joshi
183
asked Nov 14 at 6:35
Amogh Joshi
183
asked Nov 14 at 6:35
Amogh Joshi
183
183
add a comment |
add a comment |
2 Answers
2
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3
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accepted
$$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$
Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
I am getting the answer ln(7/5). Is it right?
– Amogh Joshi
Nov 14 at 6:49
yes, congratulations.
– Siong Thye Goh
Nov 14 at 6:51
add a comment |
up vote
4
down vote
Hint:
$$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$
Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$
Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$
and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$
Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
I am getting the answer ln(7/5). Is it right?
– Amogh Joshi
Nov 14 at 6:49
yes, congratulations.
– Siong Thye Goh
Nov 14 at 6:51
add a comment |
up vote
3
down vote
accepted
$$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$
Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
I am getting the answer ln(7/5). Is it right?
– Amogh Joshi
Nov 14 at 6:49
yes, congratulations.
– Siong Thye Goh
Nov 14 at 6:51
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$
Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
$$lim_{x to 1^+} [ln(x^7-1)- ln(x^5-1)]=ln left[lim_{xto 1^+} frac{x^7-1}{x^5-1} right]$$
Evaluate $lim_{xto 1^+} frac{x^7-1}{x^5-1} $ first and I think you can solve the problem.
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
answered Nov 14 at 6:41
Siong Thye Goh
93.4k1462114
93.4k1462114
I am getting the answer ln(7/5). Is it right?
– Amogh Joshi
Nov 14 at 6:49
yes, congratulations.
– Siong Thye Goh
Nov 14 at 6:51
add a comment |
I am getting the answer ln(7/5). Is it right?
– Amogh Joshi
Nov 14 at 6:49
yes, congratulations.
– Siong Thye Goh
Nov 14 at 6:51
I am getting the answer ln(7/5). Is it right?
– Amogh Joshi
Nov 14 at 6:49
I am getting the answer ln(7/5). Is it right?
– Amogh Joshi
Nov 14 at 6:49
yes, congratulations.
– Siong Thye Goh
Nov 14 at 6:51
yes, congratulations.
– Siong Thye Goh
Nov 14 at 6:51
add a comment |
up vote
4
down vote
Hint:
$$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$
Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$
Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$
and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
add a comment |
up vote
4
down vote
Hint:
$$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$
Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$
Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$
and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
add a comment |
up vote
4
down vote
up vote
4
down vote
Hint:
$$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$
Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$
Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$
and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
Hint:
$$lim_{xto 1^+}left[ln(x^7 -1) - ln(x^5 -1)right]=lnleft(lim_{xto 1^+}dfrac{x^7-1}{x^5-1}right)$$
Now use $x^n-1=(x-1)(x^{n-1}+x^{n-2}+cdots+x+1)$
Alternatively, $dfrac{x^7-1}{x^5-1}=dfrac{dfrac{x^7-1}{x-1}}{dfrac{x^5-1}{x-1}}$
and $$lim_{xto 1}dfrac{x^n-1}{x-1}=dfrac{d(x^n)}{dx}_{text{( at }x=1)}$$
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
edited Nov 14 at 6:45
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
answered Nov 14 at 6:40
lab bhattacharjee
220k15154271
220k15154271
add a comment |
add a comment |
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