Certainly! Let's solve the given limit step-by-step.
We need to find:
[tex]$
\lim _{x \rightarrow \infty} \frac{x^2 + \cos(x)}{x^2 + 2010}
$[/tex]
Step 1: Identify the dominant term in the numerator and the denominator.
As [tex]\( x \)[/tex] approaches infinity, the [tex]\( x^2 \)[/tex] term in both the numerator and the denominator will dominate because [tex]\( x^2 \)[/tex] grows much faster than [tex]\(\cos(x)\)[/tex] or the constant 2010.
Step 2: Factor out the dominant term [tex]\( x^2 \)[/tex] from both the numerator and the denominator.
[tex]$
\frac{x^2 + \cos(x)}{x^2 + 2010} = \frac{x^2(1 + \frac{\cos(x)}{x^2})}{x^2(1 + \frac{2010}{x^2})}
$[/tex]
Step 3: Simplify the expression by dividing both the numerator and the denominator by [tex]\( x^2 \)[/tex]:
[tex]$
= \frac{1 + \frac{\cos(x)}{x^2}}{1 + \frac{2010}{x^2}}
$[/tex]
Step 4: Evaluate the limit as [tex]\( x \)[/tex] approaches infinity.
As [tex]\( x \)[/tex] approaches infinity, [tex]\( \frac{\cos(x)}{x^2} \)[/tex] approaches 0 because [tex]\( \cos(x) \)[/tex] oscillates between -1 and 1, and [tex]\( x^2 \)[/tex] grows without bound. Similarly, [tex]\( \frac{2010}{x^2} \)[/tex] also approaches 0.
Therefore, we have:
[tex]$
\lim _{x \rightarrow \infty} \frac{1 + \frac{\cos(x)}{x^2}}{1 + \frac{2010}{x^2}} = \frac{1 + 0}{1 + 0} = \frac{1}{1} = 1
$[/tex]
So, the limit is:
[tex]$
\boxed{1}
$[/tex]
