$ \lim \limits_{x\to 0}{\tan({\cos({\frac{1}{x})})}+\lg({2+x}) \over \lg({4+x})} $
How does it solve? I don't have any ideas for it. Can you give some hints?
How solve this problem? Need some hints $ \lim \limits_{x\to 0}{\tan{\cos{\frac{1}{x}}}+\lg{2+x} \over \lg{4+x}} $
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calculus
real-analysis
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2As written, does not exists. The logs $\to$ constant, while the $\cos$ oscillates. – 2017-01-17
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0think about cos(1/x) as x approaches zero - what might the graph look like? – 2017-01-17
2 Answers
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Well, the limit of a quotient is the quotient of the limits, so the bottom is easy, and the only problem on top is the limit $$ \lim_{x \to 0} \tan \cos \frac{1}{x} $$ and this limit does not exist because it is the same as $$ \lim_{x \to 0} \tan \cos \frac{1}{x} = \tan \cos (\lim_{x \to 0} \frac{1}{x}) = \tan (\cos (\lim_{x \to \infty} x)) $$ and clearly $$\cos (\lim_{x \to \infty} x) = \lim_{x \to \infty} \cos x $$ isn't well defined as mentioned in the comment above. To see this we could look at some monotonically increasing and unbounded subsequences ${x_n}$, and if we find one that doesn't make the function converge then the function clearly can't converge. One such subsequence would be $x_n = n \pi$ which makes $ \cos x $ oscillate between $1$ and $-1$.
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0I had the same conclusion. Thanks a lot of – 2017-01-17
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From the source I see that you mean $ \lim_{x \to 0} \frac{\tan(\cos(\frac{1}{x}))+\lg(2+x)} {\lg(4+x)}$
Let $f(x)=\frac{\tan(\cos(\frac{1}{x}))+\lg(2+x)} {\lg(4+x)}$ and $x_n=\frac{1}{n \pi}$.
Then $x_n \to 0$ but $f(x_n)=\frac{\tan((-1)^n)+\lg(2+x_n)} {\lg(4+x_n)}$ and
Since $(\tan((-1)^n))$ diverges, $ \lim_{x \to 0} \frac{\tan(\cos(\frac{1}{x}))+\lg(2+x)} {\lg(4+x)}$ does not exist.