In this post, I’ll explain how to approach unusual problems involving concepts known as limits (these are numbers that represent what the limits of certain expressions are). What makes this problem unusual is that the exponents within the limits contain trigonometric functions:
$$ \lim_{x \to \infty} \frac{ 5x^{6 \, + \, \cos(x)} \, + \, 7} {3x^{6 \, + \, \cos(x)} \, – \, 100} = \frac{5}{3} $$
Actively engaging in Calculus concepts in unusual ways such as this can help with understanding other problems normally shown in a Calculus class.
Reducing the Problem
We first use a trick described in one of my earlier posts, but instead of dividing the numerator and denominator by \( x^3 \), we divide them by \( x^{6 \, + \, \cos(x)} \).
$$ \frac{ 5x^{6 \, + \, \cos(x)} \, + \, 7} {3x^{6 \, + \, \cos(x)} \, – \, 100} $$ $$ = \frac{ (5x^{6 \, + \, \cos(x)} \, + \, 7) / x^{6 \, + \, \cos(x)} } {(3x^{6 \, + \, \cos(x)} \, – \, 100) / x^{6 \, + \, \cos(x)} } $$ $$ = \frac{ 5x^{6 \, + \, \cos(x)}/x^{6 \, + \, \cos(x)} \, + \, 7/x^{6 \, + \, \cos(x)} } {3x^{6 \, + \, \cos(x)}/x^{6 \, + \, \cos(x)} \, – \, 100/x^{6 \, + \, \cos(x)} } $$ $$ = \frac{ 5 \, + \, 7/x^{6 \, + \, \cos(x)} } {3 \; – \, 100/x^{6 \, + \, \cos(x)} } $$
So,
$$ \lim_{x \to \infty} \frac{ 5x^{6 \, + \, \cos(x)} \, + \, 7} {3x^{6 \, + \, \cos(x)} \, – \, 10}$$
$$ = \lim_{x \to \infty} \frac{ 5 \, + \, 7/x^{6 \, + \, \cos(x)} } {3 \; – \, 100/x^{6 \, + \, \cos(x)} } . $$
Dealing with the Exponents
But what is the limit on the right-hand side equal to? To determine
$$ \lim_{x \to \infty} \frac{ 5 \, + \, 7/x^{6 \, + \, \cos(x)} } {3 \; – \, 100/x^{6 \, + \, \cos(x)} } $$
we need to look at \( x^{6 \, + \, \cos(x)} \).
Recall from this post that this is always true:
$$ -1 \leq \cos(x) \leq 1 $$
So the following is always true:
$$ 6 \, + \, \cos(x) \geq 6 \; – \, 1 = 5$$
And so,
$$ x^{6 \, + \, \cos(x)} \geq x^5 $$
As \( x \) goes to infinity, \( x^5 \) goes to infinity. So,
$$ \lim_{x \to \infty} x^{6 \, + \, \cos(x)} = \infty \,. $$
Putting it all together
Now, let’s look at
$$ \lim_{x \to \infty} \frac{ 5 \, + \, 7/x^{6 \, + \, \cos(x)} } {3 \; – \, 100/x^{6 \, + \, \cos(x)} }. $$
Let’s look at the expressions \( 7/x^{6 \, + \, \cos(x)} \) and \( 100/x^{6 \, + \, \cos(x)} \).
As \( x \) goes to infinity, \( x^{6 \, + \, \cos(x)} \) goes to infinity. So,
$$ \lim_{x \to \infty} \frac{7}{x^{6 \, + \, \cos(x)}} = 0 $$
and
$$ \lim_{x \to \infty} \frac{100}{x^{6 \, + \, \cos(x)}} = 0 $$
So,
$$ \lim_{x \to \infty} \frac{ 5 \, + \, 7/x^{6 \, + \, \cos(x)} } {3 \; – \, 100/x^{6 \, + \, \cos(x)} } = \frac{5 \, + \, 0}{3 \; – \, 0} = \frac{5}{3} $$
This is the answer we’re looking for.
