Qfyilyi

Q - $$lim_{xto 0} frac{tan x - sin x}{x^3}$$

Sol -
$$Rightarrow qquad lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$

$$Rightarrow qquad lim_{xto 0}frac{tan x}{x}.lim_{xto 0} frac{1}{x^2} -lim_{xto 0} frac{sin x}{x}.lim_{xto 0} frac{1}{x^2}$$

$$Rightarrow qquad 1.lim_{xto 0} frac{1}{x^2} -1.lim_{xto 0} frac{1}{x^2}$$

$$Rightarrow qquad 1.lim_{xto 0} frac{1}{x^2} -1.lim_{xto 0} frac{1}{x^2}$$

$$Rightarrow qquad lim_{xto 0} frac{1}{x^2} -frac{1}{x^2}$$
$$Rightarrow qquad lim_{xto 0} space 0$$
$$0$$

But the answer is $frac{1}{2}$ by L'Hopital's Rule.

Your problem arises from the fact that you used $color{red}{lim_limits{x to 0} frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $color{red}{infty-infty}$...

Only split an initial limit into a product if the individual limits are defined.

This is just another way of saying what the others told you.

$$lim_{xto 0} frac{tan x - sin x}{x^3}
ne lim_{xto 0} frac{tan x}{x^3} - lim_{xto 0} frac{sin x}{x^3}$$

The theorem is
IF $displaystyle lim_{xto 0}f(x) = L$
and $displaystyle lim_{xto 0}g(x)=M$, where $M, N in mathbb R$,
THEN $displaystyle lim_{xto 0}(f(x)-g(x))=L-M$

But, since $displaystyle lim_{xto 0} frac{tan x}{x^3} = lim_{xto 0} frac{sin x}{x^3} = infty$, then the theorem does not apply.

This limit can be evaluated without resorting to L'Hospital.

begin{align}
frac{tan x - sin x}{x^3}
&= frac{frac{sin x}{cos x} - sin x}{x^3} \
&= frac{sin x - sin x cos x}{x^3 cos x} \
&= frac{1}{cos x} cdotfrac{sin x}{x} cdot frac{1 - cos x}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac{2sin^2(frac 12x)}{x^2} \
&= frac{1}{cos x} cdotfrac{sin x}{x}
cdot frac 12 cdot left(frac{sin frac x2}{frac x2}right)^2 \
end{align}

which approaches $dfrac 12$ as $x$ approaches $0$.

I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ limlimits_{x to 0}big( f(x) - g(x)big)$ is not always equal to $ limlimits_{x to 0} f(x) - limlimits_{x to 0} g(x)$.

Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $sin$ & $tan$ as Taylor series - each has first term in $theta^1$ with coefficient 1, so in the difference it drops out. If you plot $sintheta-tantheta$ it looks like a cubic at the origin. Then if you divide that series by $theta^3$, & you get a series with an initial term in $theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $thetarightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.

To actually get the answer immediately you just subtract the coefficient for $theta^3$ in the series for $sintheta$ from that in that for $tantheta$, & you get 1/3 - -1/6 = 1/2.

Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.

We have

$$lim_{x to 0} tan x= frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . .$$

$$lim_{x to 0} sin x= frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .$$

Therefore expression turns to,

$$lim_{x to 0} frac{frac{x}{1} +frac{x^{3}}{3} +frac{2x^{5}}{15} + . . . - (frac{x}{1} - frac{x^{3}}{6} +frac{x^{5}}{120} + . . .)}{x^{3}}$$

Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $frac{1}{3}$ and $frac{1}{6}$ which is $color{red} {frac{1}{2}}$