# Help with the discussion of a limit with β\beta and γ\gamma

The discussion of the following limit:

limx−>0+(xβ.ln(x)γ)\lim_{x->0+}\left(x^\beta.\ln(x)^\gamma\right)

1) When β=0\beta=0 and γ=0\gamma=0 : limx→0x0.ln(x)0=1\lim_{x \rightarrow 0}{x^0.\ln(x)^0}=1

2) When β>0\beta>0 and γ=0\gamma=0 : limx→0xβ.ln(x)0=\lim_{x \rightarrow 0}{x^\beta.\ln(x)^0}= limx→0xβ=0\lim_{x \rightarrow 0}x^\beta=0

3) When \beta<0\beta<0 and \gamma=0\gamma=0 : \lim_{x->0}x^\beta.\ln(x)^0\lim_{x->0}x^\beta.\ln(x)^0= \lim_{x \rightarrow 0}x^\beta= +\infty \lim_{x \rightarrow 0}x^\beta= +\infty

4) \beta=0\beta=0 and \gamma>0\gamma>0 : \lim_{x \rightarrow 0}{x^0.\ln(x)^\gamma}= \lim_{x \rightarrow 0}{x^0.\ln(x)^\gamma}= \lim_{x \rightarrow 0}(\ln x)^\gamma= \infty \lim_{x \rightarrow 0}(\ln x)^\gamma= \infty

5) \beta=0\beta=0 and \gamma<0\gamma<0 : \lim_{x \rightarrow 0}{x^0.\ln(x)^\gamma}= \lim_{x \rightarrow 0}{x^0.\ln(x)^\gamma}= \lim_{x \rightarrow 0}(\ln x)^\gamma= \lim_{x \rightarrow 0} \lim_{x \rightarrow 0}(\ln x)^\gamma= \lim_{x \rightarrow 0} \frac{1}{(\ln x)^-\gamma}=0\frac{1}{(\ln x)^-\gamma}=0 now here is where im stuck: \beta>0\beta>0 and \gamma>0\gamma>0 : \lim_{x \rightarrow 0}{x^\beta.\ln(x)^\gamma}= lim_{x \rightarrow 0}\frac{(lnx)^\gamma}{\frac{1}{x^\beta}} \lim_{x \rightarrow 0}{x^\beta.\ln(x)^\gamma}= lim_{x \rightarrow 0}\frac{(lnx)^\gamma}{\frac{1}{x^\beta}}

should i discuss for \beta>\gamma>0 ?\beta>\gamma>0 ?
im confused

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