Mixing
Limit of a natural log function
Lim x-->0 ln((x^2)/16-x^2)
How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!
limits
asked Nov 20 '18 at 0:01
M Do
124
add a comment |
Lim x-->0 ln((x^2)/16-x^2)
How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!
limits
asked Nov 20 '18 at 0:01
M Do
124
4
Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05
@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54
add a comment |
Lim x-->0 ln((x^2)/16-x^2)
How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!
limits
asked Nov 20 '18 at 0:01
M Do
124
Lim x-->0 ln((x^2)/16-x^2)
How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!
limits
limits
asked Nov 20 '18 at 0:01
M Do
124
asked Nov 20 '18 at 0:01
M Do
124
asked Nov 20 '18 at 0:01
M Do
124
asked Nov 20 '18 at 0:01
M Do
124
asked Nov 20 '18 at 0:01
M Do
124
124
4
Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05
@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54
add a comment |
4
Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05
@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54
4
4
Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05
Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05
@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54
@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54
add a comment |
1 Answer
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The answer is $-infty$. We have
$$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.
Therefore, the answer is $-infty$.
answered Nov 20 '18 at 1:26
Ekesh
5326
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The answer is $-infty$. We have
$$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.
Therefore, the answer is $-infty$.
answered Nov 20 '18 at 1:26
Ekesh
5326
add a comment |
The answer is $-infty$. We have
$$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.
Therefore, the answer is $-infty$.
answered Nov 20 '18 at 1:26
Ekesh
5326
add a comment |
The answer is $-infty$. We have
$$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.
Therefore, the answer is $-infty$.
answered Nov 20 '18 at 1:26
Ekesh
5326
The answer is $-infty$. We have
$$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.
Therefore, the answer is $-infty$.
answered Nov 20 '18 at 1:26
Ekesh
5326
answered Nov 20 '18 at 1:26
Ekesh
5326
answered Nov 20 '18 at 1:26
Ekesh
5326
answered Nov 20 '18 at 1:26
Ekesh
5326
5326
add a comment |
add a comment |
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4
Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05
@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54