# How to show that a limit exisits, and evaluate the limit of trig functions as $x$ tends to point

For each of the following functions $f$, determine whether $\lim_{x\to a}f(x)$ exists, and compute the limit if it exists. In each case, justify your answers.

$$f(x)= x^2\cos\frac{1}{x} (\sin x)^4, \text{ where } a=0$$

$$f(x)= \frac{3(\tan(2x))^2}{2x^2}, \text{ where } a=0$$

I'm awful at trig questions, (I don't think I'm allowed to use L'Hôpital's rule).

• Is this an exam you are currently taking or a past exam question? – M47145 May 18 '16 at 16:45
• this is a past exam question from 2015 from a university which has the same maths syllabus as the university that I go to. My university has a policy of not using past papers to revise from. – HELP May 18 '16 at 16:47
• I think both the limits are $0$ – Archis Welankar May 18 '16 at 16:48
• if you give your method underneath, then people can verify it. – HELP May 18 '16 at 16:50
• @HELP Thanks for clarifying. It just sounded suspicious sounding at first before the edit. – M47145 May 18 '16 at 16:50

For the first one, notice that $\cos\left(\frac{1}{x}\right)\sin^4(x)$ is bounded and $\lim_{x\to0}x^2=0$.

For the second one, use the product rule and you can use Taylor series (or l'Hôpital's rule):

$$\lim_{x\to0}\frac{3\tan^2(2x)}{2x^2}=\frac{3}{2}\left(\lim_{x\to0}\frac{1}{\cos^2(2x)}\right)\left(\lim_{x\to0}\frac{\sin^2(2x)}{x^2}\right)=\frac{3}{2}\left(\lim_{x\to0}\frac{\sin^2(2x)}{x^2}\right)$$

Part (a):

$$\lim_{x\to 0}x^2\cos(1/x)(\sin x)^4$$ as $\cos(\mathrm{anything})=$ is defined, but here it is oscillating between $-1$ and $1$ so we can apply limit $$\lim_{x\to 0}x^2\cos(1/x)(\sin x)^4=(0)*n*(0)$$ where $n=$ some oscillating(but finite) number

hence answer is $0$.

Part (b):

this one is quite straight forward you might be knowing that $\dfrac{\tan x}{x}=1$ when $x\to 0$(why?you can ask in comments) then similarly $\bigg(\dfrac{\tan x}{x}\bigg)^2=1$ when $x\to 0$

so just mutliply and divide you will get$$\lim_{x\to 0}\,\,({3\,\,\text x\,\,2})\bigg(\dfrac{\tan 2x}{2x}\bigg)^2=6$$

Since the trigonometric functions are in radians, we can say that if $x>0$ then $u < \tan u$, and if $u$ is close enough to $0$, then $\tan u < 1.1u$. Thus with $u=2x$, if $x>0$, we have $$1 < \frac{\tan(2x)}{2x} < 1.1.$$ Since $u \mapsto \dfrac{\tan u} u$ is an even function, this must also hold when $u$ is negative and close enough to $0$.

Then show that what was done with $1.1$ can be done with any other number bigger than $1$ (but how close is "close enough" above depends on how close that number is to $1$).

With all this we can show that $\displaystyle\lim_{x\to0} \frac{\tan(2x)}{2x} = 1$.

But if you're terrible at trig questions, maybe hitting the trigonometry books is what you need to be able to understand this.