求x*ln(x+根号下1+x^2)-根号下1+x^2在x趋向负无穷的极限
时间: 2023-09-11 16:07:42 浏览: 53
首先,我们观察被求极限的函数:
f(x) = xln(x+√(1+x^2)) - √(1+x^2)
当x趋向负无穷时,我们可以将f(x)中的√(1+x^2)看作x的高次项,从而得到:
f(x) ≈ xln(x+x) - x
f(x) ≈ xln(2x) - x
因此,我们可以将原式重写为:
f(x) = xln(2x) - x - xln(x+√(1+x^2))
f(x) = xln(2x) - x - xln(x(1+√(1/x^2)))
f(x) = xln(2x) - x - xln(x) - xln(1+√(1/x^2))
接下来,我们分别考虑每一项的极限:
lim xln(2x) = -∞ (当x趋向负无穷时)
lim xln(x) = -∞ (当x趋向负无穷时)
lim xln(1+√(1/x^2)) = lim xln(1+1/|x|) = 0 (当x趋向负无穷时)
因此,原式的极限为:
lim xln(x+√(1+x^2)) - √(1+x^2) = -∞
注意:在上述计算中,我们使用了对数函数和根号函数的渐进性质,即:
ln(x) / x → 0 (当x趋向正无穷时)
ln(x) / x → -∞ (当x趋向负无穷时)
√(x^2+1) / x → 1 (当x趋向正无穷时)
√(x^2+1) / x → 1 (当x趋向负无穷时)
相关问题
求x乘以ln(x+根号下1+x^2)-根号下1+x^2在x趋向负无穷的极限是多少
我们可以使用洛必达法则来求解这个极限。首先,我们注意到在$x$趋向负无穷时,$\sqrt{1+x^2}$的表现类似于$x$,因此可以将原式改写为:
$$
\begin{aligned}
& x\ln(x+\sqrt{1+x^2}) - \sqrt{1+x^2} \\
=& x\ln(x+\sqrt{1+x^2}) - \sqrt{1+x^2} \cdot \frac{x}{\sqrt{1+x^2}} \cdot \frac{\ln(x+\sqrt{1+x^2})}{x} \\
=& x\left[\ln(x+\sqrt{1+x^2}) - \frac{\sqrt{1+x^2}}{x}\cdot\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\right]
\end{aligned}
$$
接下来,我们可以使用洛必达法则来求解极限。具体地,我们计算分子和分母的导数,并且分别求它们在$x$趋向负无穷的极限,如下所示:
$$
\begin{aligned}
\lim_{x\to-\infty} x\left[\ln(x+\sqrt{1+x^2}) - \frac{\sqrt{1+x^2}}{x}\cdot\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\right] &= \lim_{x\to-\infty} \frac{\ln(x+\sqrt{1+x^2}) - \frac{\sqrt{1+x^2}}{x}\cdot\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}}{\frac{1}{x}} \\
&= \lim_{x\to-\infty} \frac{\frac{1}{\sqrt{1+x^2}} - \frac{\sqrt{1+x^2}}{x^2}\cdot\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} - \frac{\ln(x+\sqrt{1+x^2})}{x\sqrt{1+x^2}}}{-\frac{1}{x^2}} \\
&= \lim_{x\to-\infty} \frac{-\frac{x}{(1+x^2)^{\frac{3}{2}}} - \frac{1}{x^2}\cdot\frac{\sqrt{1+x^2}}{x+\sqrt{1+x^2}} - \frac{\sqrt{1+x^2}}{x^2\sqrt{1+x^2}} - \frac{\ln(x+\sqrt{1+x^2})}{x^2\sqrt{1+x^2}}}{\frac{1}{x^2}} \\
&= \lim_{x\to-\infty} \frac{-x^3 - (1+x^2)\sqrt{1+x^2} - x\sqrt{1+x^2}\ln(x+\sqrt{1+x^2})}{(1+x^2)^{\frac{3}{2}}x^2} \\
&= \lim_{x\to-\infty} \frac{-x^3(1+\frac{1}{x^2}) - \sqrt{1+\frac{1}{x^2}}(1+\frac{1}{x^2})\sqrt{1+\frac{1}{x^2}} - x\sqrt{1+\frac{1}{x^2}}\ln(x+\sqrt{1+\frac{1}{x^2}})}{(1+\frac{1}{x^2})^{\frac{3}{2}}} \\
&= \lim_{x\to-\infty} \frac{-x^3 - \sqrt{1+\frac{1}{x^2}} - x\sqrt{1+\frac{1}{x^2}}\ln(x+\sqrt{1+\frac{1}{x^2}})}{(1+\frac{1}{x^2})^{\frac{3}{2}}} \\
&= \lim_{x\to-\infty} \frac{-x^3 - \sqrt{1+\frac{1}{x^2}} - x\sqrt{1+\frac{1}{x^2}}\ln(1+\sqrt{1+\frac{1}{x^2}})}{(1+\frac{1}{x^2})^{\frac{3}{2}}}
\end{aligned}
$$
最后一步是因为$\sqrt{1+\frac{1}{x^2}}-1$趋向于$0$,因此可以将$\ln(x+\sqrt{1+\frac{1}{x^2}})$近似为$\ln(1+\sqrt{1+\frac{1}{x^2}})$。
现在我们可以使用洛必达法则来计算极限了。具体地,我们计算分子和分母的导数,并且分别求它们在$x$趋向负无穷的极限,如下所示:
$$
\begin{aligned}
\lim_{x\to-\infty} \frac{-x^3 - \sqrt{1+\frac{1}{x^2}} - x\sqrt{1+\frac{1}{x^2}}\ln(1+\sqrt{1+\frac{1}{x^2}})}{(1+\frac{1}{x^2})^{\frac{3}{2}}} &= \lim_{x\to-\infty} \frac{-3x^2 - \frac{1}{x^3} + \frac{\sqrt{1+\frac{1}{x^2}}}{x^2} - \frac{1}{\sqrt{1+\frac{1}{x^2}}} - \frac{\sqrt{1+\frac{1}{x^2}}}{x^2}\ln(1+\sqrt{1+\frac{1}{x^2}}) - \frac{1}{x^2(1+\sqrt{1+\frac{1}{x^2}})}}{\frac{3}{2}(1+\frac{1}{x^2})^{\frac{1}{2}}\cdot\frac{-1}{x^3}} \\
&= \lim_{x\to-\infty} \frac{6x^5+2x+\frac{3x^2+1}{x^5}\sqrt{1+\frac{1}{x^2}} - \frac{3x^2+1}{x^5}\ln(1+\sqrt{1+\frac{1}{x^2}}) - \frac{2}{x^3(1+\sqrt{1+\frac{1}{x^2}})}}{-3(1+\frac{1}{x^2})^{\frac{1}{2}}} \\
&= \lim_{x\to-\infty} \frac{6x^5+2x-\frac{6x^3+2x}{x^5}\ln(1+\sqrt{1+\frac{1}{x^2}})}{-3(1+\frac{1}{x^2})^{\frac{1}{2}}} \\
&= \lim_{x\to-\infty} \frac{-6x^5-2x}{3(1+\frac{1}{x^2})^{\frac{1}{2}}} \\
&= -\infty
\end{aligned}
$$
因此,原式在$x$趋向负无穷的极限是$-\infty$。
求1+x乘以ln(x+根号下1+x^2)-根号下1+x^2在x趋向负无穷的极限
我们可以利用洛必达法则来求解这个极限。首先我们需要将原式化简:
\begin{aligned} &1+x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}\\ =&\frac{(1+x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2})(1+x\ln(x+\sqrt{1+x^2})+\sqrt{1+x^2})}{1+x\ln(x+\sqrt{1+x^2})+\sqrt{1+x^2}}\\ =&\frac{(1+x\ln(x+\sqrt{1+x^2}))^2-(1+x^2)}{1+x\ln(x+\sqrt{1+x^2})+\sqrt{1+x^2}}\\ \end{aligned}
因此,我们需要计算的极限就是:
\begin{aligned} \lim_{x\to-\infty}\frac{(1+x\ln(x+\sqrt{1+x^2}))^2-(1+x^2)}{1+x\ln(x+\sqrt{1+x^2})+\sqrt{1+x^2}} \end{aligned}
对上式分子和分母同时除以 $x^2$,得到:
\begin{aligned} \lim_{x\to-\infty}\frac{(\frac{1}{x}+\ln(x+\sqrt{1+x^2}))^2-1}{\frac{1}{x}+\ln(x+\sqrt{1+x^2})+\sqrt{1+\frac{1}{x^2}}} \end{aligned}
由于 $x\to-\infty$,因此 $\frac{1}{x}\to0$。因此,上式的极限可以转化为:
\begin{aligned} \lim_{x\to-\infty}\frac{(\frac{1}{x}+\ln(x+\sqrt{1+x^2}))^2-1}{\frac{1}{x}+\ln(x+\sqrt{1+x^2})} \end{aligned}
对上式使用洛必达法则,得到:
\begin{aligned} \lim_{x\to-\infty}\frac{2(\frac{1}{x}+\ln(x+\sqrt{1+x^2}))(1+\frac{x}{\sqrt{1+x^2}})-\frac{1}{x^2}}{1+\frac{1}{x+\sqrt{1+x^2}}} \end{aligned}
再次对上式使用洛必达法则,得到:
\begin{aligned} \lim_{x\to-\infty}\frac{2(\frac{1}{x}+\ln(x+\sqrt{1+x^2}))(1+\frac{x}{\sqrt{1+x^2}})-\frac{1}{x^2}}{1+\frac{1}{x+\sqrt{1+x^2}}}=&\lim_{x\to-\infty}\frac{(2+\frac{1}{\sqrt{1+x^2}})(1+\frac{\sqrt{1+x^2}}{x}-\frac{1}{x^2\sqrt{1+x^2}})}{1+\frac{1}{x+\sqrt{1+x^2}}}\\ =&\frac{2}{1-0}=\boxed{2} \end{aligned}