Find the limit

Take the natural log of the expression to get $x*\ln\left(\cos\left(\dfrac{1}{x}\right)+\dfrac{1}{x}\right)$ .

Now let $y=\dfrac{1}{x}$ and the expression becomes $\dfrac{\ln\left(\cos(y)+y\right)}{y}$ .

$\displaystyle \lim_{y \to 0} \left(\dfrac{\ln\left(\cos(y)+y\right)}{y}\right)$ satisfies the $\dfrac{0}{0}$ condition for l'Hopital's Rule, so we may evaluate by taking the derivatives of numerator and denominator, which gives $\displaystyle \lim_{y \to 0} \left(\dfrac{1-\sin(y)}{\cos(y)+y}\right)$ .

This limit may be directly evaluated, and its result is $1$ .

Finally, we recall the natural logarithm taken at the first step. $1$ is the limit of the natural logarithm of the expression, so $e^1=e\approx\boxed{2.718}$ is the limit of the original expression itself.

$\displaystyle \lim_{x \to \infty}{\cos(\dfrac{1}{x})=1}$ . So the limit is equivalent to $\displaystyle \lim_{x \to \infty}{\left(1+\dfrac{1}{x} \right)}^x=e \approx 2.718$

Let the limit be $L$ , then we have:

$\begin{aligned} L & = \lim_{x \to \infty} \left(\color{#3D99F6}{\cos\left(\frac{1}{x}\right)} + \frac{1}{x} \right)^x \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \lim_{x \to \infty} \left(\color{#3D99F6}{\left( 1 - \frac{1}{2!x^2} + \frac{1}{4!x^4} -... \right)} + \frac{1}{x} \right)^x \quad \quad \small \color{#3D99F6}{\text{Since, when } x \to \infty, \frac{1}{x} \gg \frac{1}{x^2} \gg \frac{1}{x^4} \gg \frac{1}{x^6} \gg ...} \\ & = \lim_{x \to \infty} \left(1+\frac{\color{#3D99F6}{1}}{x} \right)^x = e^\color{#3D99F6}{1} \approx \boxed{2.718} \end{aligned}$

$\lim _{ x->\infty }{ { \left( \cos { \frac { 1 }{ x } } +\frac { 1 }{ x } \right) }^{ x } } =\lim _{ y->0 }{ { \left( \cos { y } +y \right) }^{ \frac { 1 }{ y } } } \quad where\quad \frac { 1 }{ x } =y\\ =\lim _{ y->0 }{ { \left( \cos { y } +y+1-1 \right) }^{ (\frac { 1 }{ \cos { y+y-1 } } )*\frac { cosy+y-1 }{ y } } } \\ using\quad \lim _{ x->0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e\\ the\quad limit\quad transforms\quad to\quad { e }^{ \lim _{ y->0 }{ \frac { \left( \cos { y+y-1 } \right) }{ y } } }\\ As\quad both\quad the\quad functions\quad are\quad continous\quad and\quad differentiable\\ Using\quad L'Hopital's\quad rule\quad the\quad limit\quad transforms\quad to\quad \\ { e }^{ \lim _{ y->0 }{ \frac { siny+1 }{ 1 } } }={ e }^{ 1 }\approx 2.718$