How do you use the epsilon delta definition to find the limit of #x^2 cos(1/x)# as x approaches #0#?

Question

Cameron Alexander · Accepted Answer

#lim_(x->0) x^2cos(1/x) = 0#

The epsilon-delta formulation of the definition of limit states that:

#lim_(x->0) f(x) = 0#

if:

#AA epsilon >0, EE delta >0# such that for #|x| < delta, |f(x)| < epsilon#

So let's look at #f(x)# around #x=0# and note that:

#|x^2cos(1/x)|<=|x^2||cos(1/x)| <= |x^2|#

as #cosx# is bounded in the range #[-1,1]#.

So, given any #epsilon >0# take #delta < sqrt(epsilon# Then for #x in (-delta, delta)#, of course #|x| < delta#, so that:

#|x^2cos(1/x)| <= |x^2| < delta^2 < epsilon#

which proves that:

#lim_(x->0) x^2cos(1/x) = 0#

Christopher Agresta · Answer

undefined To use the epsilon-delta definition to find the limit of $x^2 \cos(1/x)$ as $x$ approaches 0, we need to show that for any given epsilon ($\varepsilon$) greater than 0, there exists a delta ($\delta$) greater than 0 such that if $0 < |x - 0| < \delta$, then $|x^2 \cos(1/x) - L| < \varepsilon$, where $L$ is the limit we are trying to find.

First, let's simplify the expression $x^2 \cos(1/x)$. Since the cosine function is bounded between -1 and 1, we can say that $|x^2 \cos(1/x)| \leq |x^2|$ for all $x$. Therefore, we can focus on finding the limit of $x^2$ as $x$ approaches 0.

Now, let's consider the inequality $|x^2 - L| < \varepsilon$. We want to find a suitable $\delta$ such that if $0 < |x - 0| < \delta$, then $|x^2 - L| < \varepsilon$.

To do this, we can manipulate the inequality as follows:

$|x^2 - L| < \varepsilon$
$-\varepsilon < x^2 - L < \varepsilon$
$L - \varepsilon < x^2 < L + \varepsilon$

Since we are interested in the limit as $x$ approaches 0, we can assume that $|x|$ is small. Therefore, we can choose $\delta$ such that $\delta < 1$ (for example).

Now, if we choose $\delta = \sqrt{L + \varepsilon} - \sqrt{L - \varepsilon}$, we can show that if $0 < |x - 0| < \delta$, then $|x^2 - L| < \varepsilon$.

By substituting $\delta$ into the inequality, we get:

$L - \varepsilon < x^2 < L + \varepsilon$
$L - \varepsilon < x^2 < L + \varepsilon$
$L - \varepsilon < (\sqrt{L + \varepsilon} - \sqrt{L - \varepsilon})^2 < L + \varepsilon$

Expanding the square and simplifying, we have:

$L - \varepsilon < L + \varepsilon - 2\sqrt{(L + \varepsilon)(L - \varepsilon)} < L + \varepsilon$

Since $\delta < 1$, we can conclude that if $0 < |x - 0| < \delta$, then $|x^2 - L| < \varepsilon$.

Therefore, the limit of $x^2 \cos(1/x)$ as $x$ approaches 0 is $L$.