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Limit of y=lnx
- Thread starter fred2028
- Start date
\(\displaystyle \lim_{x\to{0^{+}}}ln(x) = {-\infty}\)
If we let \(\displaystyle v=\frac{1}{x}\), then \(\displaystyle v\to{\infty}\) as \(\displaystyle {x\to{0^{+}}}\).
Then we get \(\displaystyle \lim_{x\to{0^{+}}}ln(x) = \lim_{v\to{\infty}}ln(\frac{1}{v}) = \lim_{v\to{\infty}}(-ln(v)) = -\lim_{v\to{\infty}}ln(v) = {-\infty}\)
If we let \(\displaystyle v=\frac{1}{x}\), then \(\displaystyle v\to{\infty}\) as \(\displaystyle {x\to{0^{+}}}\).
Then we get \(\displaystyle \lim_{x\to{0^{+}}}ln(x) = \lim_{v\to{\infty}}ln(\frac{1}{v}) = \lim_{v\to{\infty}}(-ln(v)) = -\lim_{v\to{\infty}}ln(v) = {-\infty}\)
Hmm yes but in your work how did you know that the limit of ln(v) as v -> infinity is infinity, apart from the graph?galactus said:\(\displaystyle \lim_{x\to{0^{+}}}ln(x) = {-\infty}\)
If we let \(\displaystyle v=\frac{1}{x}\), then \(\displaystyle v\to{\infty}\) as \(\displaystyle {x\to{0^{+}}}\).
Then we get \(\displaystyle \lim_{x\to{0^{+}}}ln(x) = \lim_{v\to{\infty}}ln(\frac{1}{v}) = \lim_{v\to{\infty}}(-ln(v)) = -\lim_{v\to{\infty}}ln(v) = {-\infty}\)