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Limits

lim⁡x→0+(x+1)ln⁡xsin⁡x\displaystyle \lim_{x \to 0^{+}} \frac{(x+1)\ln{x}}{\sin{x}}

I don't know what to do with this one. :eek:
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math42
20
Original post by Whistlejacket
lim⁡x→0+(x+1)ln⁡xsin⁡x\displaystyle \lim_{x \to 0^{+}} \frac{(x+1)\ln{x}}{\sin{x}}

I don't know what to do with this one. :eek:


As x -> 0, sinx can be approximated by x?
Original post by 1 8 13 20 42
As x -> 0, sinx can be approximated by x?
Thanks.

So lim⁡x→0+(x+1)ln⁡xsin⁡x=lim⁡x→0+(x+1)ln⁡xx=1+lim⁡x→0+ln⁡xx \displaystyle \lim_{x \to 0^{+}} \frac{(x+1)\ln{x}}{\sin{x}} =\lim_{x \to 0^{+}} \frac{(x+1)\ln{x}}{x} = 1+\lim_{x \to 0^{+}} \frac{\ln{x}}{x}

lim⁡x→0+ln⁡xx=lim⁡x→0+1xlim⁡x→0+ln⁡x=∞⋅−∞=−∞.\displaystyle \lim_{x \to 0^{+}} \frac{\ln{x}}{x} = \lim_{x \to 0^+}\frac{1}{x}\lim_{x \to 0^+}\ln{x} = \infty \cdot -\infty = -\infty.

Is that correct?
Reply 3
Avatar for math42
math42
20
Original post by Whistlejacket
Thanks.

So lim⁡x→0+(x+1)ln⁡xsin⁡x=lim⁡x→0+(x+1)ln⁡xx=1+lim⁡x→0+ln⁡xx \displaystyle \lim_{x \to 0^{+}} \frac{(x+1)\ln{x}}{\sin{x}} =\lim_{x \to 0^{+}} \frac{(x+1)\ln{x}}{x} = 1+\lim_{x \to 0^{+}} \frac{\ln{x}}{x}

lim⁡x→0+ln⁡xx=lim⁡x→0+1xlim⁡x→0+ln⁡x=∞⋅−∞=−∞.\displaystyle \lim_{x \to 0^{+}} \frac{\ln{x}}{x} = \lim_{x \to 0^+}\frac{1}{x}\lim_{x \to 0^+}\ln{x} = \infty \cdot -\infty = -\infty.

Is that correct?


Answer should be correct but I'm not sure where the 1 comes from. you have lim (1 + 1/x)(lnx) which should give the answer immediately?
Original post by 1 8 13 20 42
Answer should be correct but I'm not sure where the 1 comes from. you have lim (1 + 1/x)(lnx) which should give the answer immediately?
You're right, I got the 1 from nowhere. It's basic algebra failure. The answer does fall out immediately, thanks again!

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