I need help solving this problem. I tried L'hospital and rearranging but nothing worked. $$ \lim_\limits{x→∞} x^2\left(\ln\left(1 + \frac{1}{x}\right)- \frac{1}{x+1}\right) $$
Limit of function with natural logarithm
1
$\begingroup$
limits
logarithms
4 Answers
0
We have
$$x(\frac{\ln (1+\frac{1}{x})}{\frac{1}{x}})-\frac{x^2}{1+x}$$
Let $x=\frac{1}{y}$. We have,
$$=\frac{1}{y}\frac{ \ln (1+y)}{y}-\frac{1}{y^2+y}$$
This is,
$$=\frac{1}{y} \frac{ \ln (1+y)}{y}-\frac{1}{y}+\frac{1}{y+1}$$
$$=\frac{1}{y}(\frac{\ln(1+y)}{y}-1)+\frac{1}{y+1}$$
$$=\frac{\ln(1+y)-y}{y^2}+\frac{1}{y+1}$$
The first limit, as $y \to 0^+$ can be computed by LH used twice.
$$ \to -\frac{1}{2}+1=\frac{1}{2}$$
1
write your limit in the form $$\lim_{x \to \infty}\frac{\ln\left(1+\frac{1}{x}\right)-\frac{1}{x+1}}{\frac{1}{x^2}}$$ and use L'Hospital
1
Using L'Hospital rule we get $$\\ \lim _{ x\rightarrow \infty }{ \frac { \ln { \left( 1+\frac { 1 }{ x } \right) -\frac { 1 }{ x+1 } } }{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow \infty }{ \frac { \frac { x }{ x+1 } \cdot \left( -\frac { 1 }{ { x }^{ 2 } } \right) +\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } }{ \frac { -2 }{ { x }^{ 3 } } } } =\\ =\lim _{ x\rightarrow \infty }{ \frac { -x-1+x }{ { x\left( x+1 \right) }^{ 2 } } \cdot \frac { { x }^{ 3 } }{ -2 } } =\lim _{ x\rightarrow \infty }{ \frac { 1 }{ { \left( x+1 \right) }^{ 2 } } \cdot \frac { { x }^{ 2 } }{ 2 } } =\frac { 1 }{ 2 } $$
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Here is how I would do it quickly. Note that most of these steps could be done mentally, but I have written them out for the OP. $$\lim_{x\to \infty} x^2\left(\log(1 + 1/x)- \frac{1}{x+1}\right)$$ We let $u=1/x$ and multiplying the rightmost fraction by $u$ on top and bottom $$\lim_{u\to 0^+} \frac{\ln (1+u)-\frac{u}{u+1}}{u^2}$$ $$\lim_{u\to 0^+}\frac{\left(u+1\right)\ln (1+u)-u}{u^2\left(u+1\right)}$$ By Taylor Series $$\lim_{u\to 0^+}\frac{\left(u+1\right)(u-u^2/2)-u}{u^2\left(u+1\right)}$$ $$\lim_{u\to 0^+}\frac{u^2(u/2-1/2)-u}{u^2\left(u+1\right)}$$ Ignoring the lower order $u$ we get $$\lim_{u\to 0^+}\frac{u^2(u/2-1/2)}{u^2\left(u+1\right)}$$ $$\lim_{u\to 0^+}\frac{u/2-1/2}{u+1}$$ By direct evaluation we now get $$1/2$$