The limit of the subtraction of the reciprocal of sine square of angle $x$ from the reciprocal of $x$ square should be evaluated by using the formulas in this limit problem as the value of $x$ approaches zero.

It is a toughest limit problem and let’s learn how to find the limit of the reciprocal of $x$ square minus the reciprocal of sine squared of angle $x$ as the value of $x$ approaches zero by using the combination of both limit rules and some math formulas.

Evaluating the limit by the direct substitution method is failed due to the indeterminate form.

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{1}{x^2}-\dfrac{1}{\sin^2{x}}\bigg)}$ $\,=\,$ $\infty-\infty$

So, the given function is simplified by the subtraction of the fractions.

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{1}{x^2}-\dfrac{1}{\sin^2{x}}\bigg)}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^2{x}-x^2}{x^2\sin^2{x}}}$

Firstly, learn why the limit is indeterminate and how one fraction is subtracted from another fraction.

Due to the indeterminate form, the l’hospital’s rule can be used to find the limit but you learn here how to find the limit of one divided by $x$ square minus one divided by sine square of $x$ without using the l’hôpital’s rule, as the value of $x$ approaches zero.

The sine squared of angle $x$ minus $x$ square expresses the difference of squares in the numerator of the rational function and it can be factorized by the difference of squares algebraic identity.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}+x)(\sin{x}-x)}{x^2\sin^2{x}}}$

Look at the first factor in the numerator of the rational function.

- The first term is the sine of angle $x$. If the function $\sin{x}$ has $x$ in the denominator, then the limit of $\sin{x}$ divided by $x$ can be evaluated by the trigonometric limit rule in sine function.
- The second term is $x$ and it can be cancelled by itself, if it has $x$ in the denominator.

The above two factors have cleared that the terms in the first factor of the numerator require $x$ in the denominator and it can be taken from the denominator.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}+x)(\sin{x}-x)}{x^2 \times \sin^2{x}}}$

The factor $x$ square can be written as a product of $x$ and $x$ to supply one $x$ to the expression in the first factor of the rational function as the denominator.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}+x) \times (\sin{x}-x)}{x \times x \times \sin^2{x}}}$

The product of the factors in both numerator and denominator of rational function can be split as the product of two fractions.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x \times \sin^2{x}}\bigg)$

Similarly, the second factor can also be split as the product of two fractions.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x) \times 1}{x \times \sin^2{x}}\bigg)$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{\sin^2{x}}\bigg)$

Now, look at the third factor in which there is a sine function in $x$. The sine function is in square form and its limit can be evaluated, if the sine square of angle $x$ has $x$ in square form in its denominator. It is time to make the sine square of angle $x$ has $x$ square in its denominator.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{1 \times \sin^2{x}}\bigg)$

Expressing the one as the quotient of $x$ square divided by $x$ square makes the sine square of $x$ to have $x$ square in the denominator.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{\dfrac{x^2}{x^2} \times \sin^2{x}}\bigg)$

In the denominator of the third factor, the first factor is a fraction but the second factor is not a fraction. So, write the second factor as a fraction to multiply them.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{\dfrac{x^2}{x^2} \times \dfrac{\sin^2{x}}{1}}\bigg)$

The two fractions can be now multiplied in the denominator of the third factor.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{\dfrac{x^2 \times \sin^2{x}}{x^2 \times 1}}\bigg)$

The places of $x$ square and one can be changed in the third factor as per the commutative property of the multiplication.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{\dfrac{x^2 \times \sin^2{x}}{1 \times x^2}}\bigg)$

Now, the denominator in the third factor can be split as the product of two fractions.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{\dfrac{x^2}{1} \times \dfrac{\sin^2{x}}{x^2}}\bigg)$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{x^2 \times \dfrac{\sin^2{x}}{x^2}}\bigg)$

Similarly, the third factor can also be split as the product of two fractions.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1 \times 1}{x^2 \times \dfrac{\sin^2{x}}{x^2}}\bigg)$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x}$ $\times$ $\dfrac{1}{x^2}$ $\times$ $\dfrac{1}{\dfrac{\sin^2{x}}{x^2}}\bigg)$

Now, multiply the second and third factors to find their product mathematically.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x) \times 1}{x \times x^2}$ $\times$ $\dfrac{1}{\dfrac{\sin^2{x}}{x^2}}\bigg)$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x^1 \times x^2}$ $\times$ $\dfrac{1}{\dfrac{\sin^2{x}}{x^2}}\bigg)$

Look at the denominator of the second factor. Two exponents 1 and 2 having same base are involved in multiplication and their product can be evaluated by the product rule of the exponents having same bases.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x^{1+2}}$ $\times$ $\dfrac{1}{\dfrac{\sin^2{x}}{x^2}}\bigg)$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{(\sin{x}-x)}{x^3}$ $\times$ $\dfrac{1}{\dfrac{\sin^2{x}}{x^2}}\bigg)$

Use the commutative property of the multiplication to change the places of the second and third factors.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(\sin{x}+x)}{x}}$ $\times$ $\dfrac{1}{\dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{(\sin{x}-x)}{x^3}\bigg)$

According to the product rule of the limits, the limit of the product of three functions can be evaluated by the product of their limits.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}+x)}{x}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1}{\dfrac{\sin^2{x}}{x^2}}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

Look at the first factor in the expression. There is a rational function in which two functions are added in the numerator, and the rational function can be split as the sum of two like fractions.

Look at the second factor in the expression. There is a reciprocal function in which the sine function is in square form and it can be written as the square of sine function as per the trigonometry.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{\sin{x}}{x}+\dfrac{x}{x}\bigg)}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1}{\dfrac{(\sin{x})^2}{x^2}}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

In the first factor, two fractional functions are connected by a plus sign. According to the addition rule of the limits, the limit of sum of them can be evaluated by the sum of their limits. In the second factor, the expressions in denominator have the same exponent $2$ and their quotient can be evaluated by the power of a quotient rule.

$=\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$ $+$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{x}{x}\bigg)}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1}{\Big(\dfrac{\sin{x}}{x}\Big)^2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

The expression in the numerator and denominator are same in the second term of the first factor of the expression. So, the expression in the numerator $x$ gets cancelled by the expression in the denominator $x$. In the second factor, the limit of the reciprocal of a function can be evaluated by the reciprocal of its limit as per the reciprocal rule of the limits.

$=\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$ $+$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cancel{x}}{\cancel{x}}\bigg)}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{\sin{x}}{x}\Big)^2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

In the second factor, the limit of a power function can be evaluated by the power rule of the limits.

$=\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}}$ $+$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize (1)\bigg)}$ $\times$ $\dfrac{1}{\Big(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}}{x}\Big)^2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

There is the limit of sine of angle $x$ divided by $x$ as $x$ approaches zero in both the first term of the first factor and the denominator in the second factor. According to the trigonometric limit rule in sine function, the limit of sine of angle $x$ divided by $x$ is equal to one as the value of $x$ approaches zero.

Look at the second term in the first factor, the limit of one as the value of $x$ tends to zero is equal to one, as per the limit rule of a constant.

$=\,\,$ $(1+1)$ $\times$ $\dfrac{1}{(1)^2}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

Add the number $1$ to another one in the first factor of the expression to find their sum.

$=\,\,$ $(2)$ $\times$ $\dfrac{1}{1^2}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

Multiply the number $1$ by another $1$ in the denominator of fraction in the second factor of the expression to find their product.

$=\,\,$ $2$ $\times$ $\dfrac{1}{1 \times 1}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

$=\,\,$ $2$ $\times$ $\dfrac{1}{1}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

$=\,\,$ $2 \times 1$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

The two factors $2$ and $1$ can be multiplied in the expression to find their product.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(\sin{x}-x)}{x^3}}$

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{x}-x}{x^3}}$

In the numerator, the sine of angle $x$ can be written as a product of one and sine of angle $x$. Similarly, the negative $x$ can be written as a product of negative one and $x$.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1 \times \sin{x}+(-1) \times x}{x^3}}$

The number can be written as the product of two negative ones in the first term of the numerator.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(-1) \times (-1) \times \sin{x}+(-1) \times x}{x^3}}$

The negative one is a common factor in the both terms of the numerator. So, the common factor the negative one can be taken out from them to factorize the expression.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(-1) \times \big((-1) \times \sin{x}+x\big)}{x^3}}$

Now, multiply the negative one by the trigonometric function sine of angle $x$ in the numerator.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(-1) \times (-\sin{x}+x)}{x^3}}$

The places of the two terms in the second factor of the numerator can be changed by the commutative property of the addition.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(-1) \times (x-\sin{x})}{x^3}}$

The negative one is a constant and it can be released from the limit operation. It is possible when the function in fraction form is split as the product of two fractions.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{(-1) \times (x-\sin{x})}{1 \times x^3}}$

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(-1) \times (x-\sin{x})}{1 \times x^3}\bigg)}$

Now, the function in fraction form can be split as the product of two different fractions.

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{(-1)}{1} \times \dfrac{x-\sin{x}}{x^3}\bigg)}$

$=\,\,$ $2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg((-1) \times \dfrac{x-\sin{x}}{x^3}\bigg)}$

The negative one is a constant and it can be released from the limits operation by the constant multiple rule of the limits.

$=\,\,$ $2$ $\times$ $(-1) \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$

Now, the factors two and negative one can be multiplied to find their product.

$=\,\,$ $(-2)$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$

We have evaluated in another problem that the limit of $x$ minus the sine of angle $x$ divided by $x$ cube is the quotient of one divided by six as the value of $x$ approaches zero.

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6}$

Now, the limit of $x$ minus sine of angle $x$ divided by $x$ cube as the value $x$ tends to zero can be replaced by the quotient one divided by six.

$\implies$ $(-2)$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $(-2)$ $\times$ $\dfrac{1}{6}$

Now, let’s multiply the number negative two by the quotient of one divided by six to find the limit. The negative two is not a fraction but the second factor is a fraction. So, let’s write the negative two in fraction form.

$=\,\,$ $\dfrac{(-2)}{1}$ $\times$ $\dfrac{1}{6}$

Use the multiplication of the fractions to find the product of the two fractions.

$=\,\,$ $\dfrac{(-2) \times 1}{1 \times 6}$

$=\,\,$ $\dfrac{(-2)}{6}$

The negative two can be written as a product of negative one and two in the numerator.

$=\,\,$ $\dfrac{(-1) \times 2}{6}$

Similarly, the number six can be written as the product of two factors two and three in the denominator.

$=\,\,$ $\dfrac{(-1) \times 2}{2 \times 3}$

$=\,\,$ $\dfrac{(-1) \times \cancel{2}}{\cancel{2} \times 3}$

$=\,\,$ $\dfrac{(-1)}{3}$

$=\,\,$ $-\dfrac{1}{3}$

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