Yes, your answer is correct.

[navneet9431]

<Moderator's note: Moved from a technical forum and thus no template.>
$$\lim_{x\rightarrow 0} (x-tanx)/x^3$$
I solve it like this,
$$\lim_{x\rightarrow 0}1/x^2 - tanx/x^3=\lim_{x\rightarrow 0}1/x^2 - tanx/x*1/x^2$$
Now using the property $$\lim_{x\rightarrow 0}tanx/x=1$$,we have ;
$$\lim_{x\rightarrow 0}1/x^2 - 1/x^2=0$$

Is my answer correct?

 

[fresh_42]

$$\lim_{x\rightarrow 0} (x-tanx)/x^3$$
I solve it like this,
$$\lim_{x\rightarrow 0}1/x^2 - tanx/x^3=\lim_{x\rightarrow 0}1/x^2 - tanx/x*1/x^2$$
Now using the property $$\lim_{x\rightarrow 0}tanx/x=1$$,we have ;
$$\lim_{x\rightarrow 0}1/x^2 - 1/x^2=0$$

Is my answer correct?

No, it is wrong. You cannot calculate with undetermined expressions like $\frac{1}{x^2}$ as if they were numbers.
The shortest way is possibly a Taylor expansion of the function. Which theorems are you supposed to use?

 

[Delta2]

No you are wrong, you have to be carefull when you do algebra with limits, you cannot go back and forth the way you do

Instead use L'Hopital once and you ll calculate the correct value which is $\frac{-1}{3}$

 

[haruspex]

$$\lim_{x\rightarrow 0}tanx/x=1$$,we have ;
$$\lim_{x\rightarrow 0}1/x^2 - 1/x^2=0$$

To be a bit more exact regarding your error, you cannot in general turn lim(a+bc) into lim(a+c lim(b)).
The reason is that in taking the inner limit you may discard a second order term, but after multiplying what remains by c the result may then cancel with a, so the term discarded was important after all.
You may be able to rescue your method by keeping more terms and using the O() notation.

 

[navneet9431]

you may discard a second order term

About what second order term are you talking?
Note: English is my second language.
I will be thankful for any help!

 

[haruspex]

About what second order term are you talking?
Note: English is my second language.
I will be thankful for any help!

More general than saying
$$\lim_{x\rightarrow 0}tanx/x=1$$,
you can write tan(x)/x=1+x²/3+O(x⁴).
This way, after some cancellation, you have a nonzero leading term.

 

[Delta2]

When all the limits involved are finite, then the algebra with limits is pretty much the same as the algebra with real numbers.

However when some limits are infinite then the algebra of limits might break down, and in this example the limits $lim\frac{1}{x^2}$ and $\lim-\frac{1}{x^2}$ which one has limit $+\infty$ and the other $-\infty$ (assuming we take limits $x\to 0+$) and they cannot be recombined as the last step at the OP does, the rule for the sum of limits breaks down when one limit is $+\infty$ and the other is $-\infty$.