Computing the limit of an expression

Question

I have the following question. Determine the limit of the following expression. $\lim_{x \rightarrow 0} \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4}$. My attempt to this question is the following. Let the function $f(x) = \frac{cos(x \sqrt{2}) - \frac{1}{1+x^2}}{x^4} = \frac{\cos(x \sqrt{2})}{x^4} - \frac{1}{(1+x^2) x^4}.$ Then $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos(x \sqrt{2})}{x^4} - \lim_{x \rightarrow 0} \frac{1}{(1+x^2) x^4} = 0 - Undefined$. So the limit does not exists. But my professor's answer is $-\frac{5}{6}$. He used Taylor expansion. What am I doing wrong??

Answer 1

we get by simplifying your term $\frac{(1+x^2)\cos(x\sqrt{2})-1)}{x^4(1+x^2)}$ and the taylor expansion for $x=0$ is $-\frac{5}{6}+O(x^2)$

Answer 2

We have : $$\mathop {\lim }\limits_{x \to {x_0}} \left( {f(x) + g(x)} \right) = \mathop {\lim }\limits_{x \to {x_0}} f(x) + \mathop {\lim }\limits_{x \to {x_0}} g(x) ~~~~ \left(*\right) $$ ONLY IF:

$$\mathop {\lim }\limits_{x \to {x_0}} f(x) ~ and \mathop {\lim }\limits_{x \to {x_0}} g(x) ~ exist $$

Back to this problem, $\mathop {\lim }\limits_{x \to 0} {{\cos \left( {x\sqrt 2 } \right)} \over {{x^4}}} \to \infty , \mathop {\lim }\limits_{x \to 0} {1 \over {(1 + {x^2}){x^4}}}$

So $\left(*\right)$ is invalid here.

Using the Taylor Expansion: $$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{\cos \left( {x\sqrt 2 } \right) - {1 \over {1 + {x^2}}}} \over {{x^4}}} \cr & = \mathop {\lim }\limits_{x \to 0} {{(1 + {x^2})\cos \left( {x\sqrt 2 } \right) - 1} \over {{x^4}(1 + {x^2})}} \cr & = \mathop {\lim }\limits_{x \to 0} {{(1 + {x^2})\left( {1 - {1 \over 2}{{\left( {x\sqrt 2 } \right)}^2} + {1 \over {24}}{{\left( {x\sqrt 2 } \right)}^4} + o\left( {{x^4}} \right)} \right) - 1} \over {{x^4}}} \cr & = \mathop {\lim }\limits_{x \to 0} {{ - {5 \over 6}{x^4} + o\left( {{x^4}} \right)} \over {{x^4}}} \cr & = - {5 \over 6} \cr} $$