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

The limit of the one divided by $x$ square minus one divided by sine square of $x$ should be evaluated in this limit problem as the value of $x$ approaches zero.

Find the Limit by the Direct substitution

Firstly, let’s try the direct substitution method to find the limit as the value of $x$ tends to zero.

$=\,\,$ $\dfrac{1}{0^2}-\dfrac{1}{\sin^2{(0)}}$

The square of the quantities can be written in product form as per the exponentiation.

$=\,\,$ $\dfrac{1}{0 \times 0}-\dfrac{1}{\sin{(0)} \times \sin{(0)}}$

According to trigonometry, the sine of zero radian is equal to zero. So, substitute it in the second term of the expression.

$=\,\,$ $\dfrac{1}{0 \times 0}-\dfrac{1}{0 \times 0}$

Now, calculate the product in the denominator of each term by multiplication.

$=\,\,$ $\dfrac{1}{0}-\dfrac{1}{0}$

The reciprocal of zero is infinity as per the fundamental mathematics.

$=\,\,$ $\infty-\infty$

It is evaluated that the limit of the reciprocal of $x$ square minus the reciprocal of sine square of angle $x$ is indeterminate as the value of $x$ is closer to zero. It expresses that the direct substitution method is not suitable to find the limit of this function. So, we should think about alternative methods.

Find the difference by subtracting fractions

The one divided by $x$ square and the one divided by square of sine of angle $x$ are the unlike fractions. So, the difference between them cannot be evaluated directly but it can be done by making both fractions to have the same denominators.

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

The expressions in the denominators are the $x$ square and the sine squared of angle $x$, and they do not have common factors.

1. In the first term, the fraction should have sine square of angle $x$ in the denominator. So, the factor $1$ in the first term can be written as the quotient of the sine square of angle $x$ divided by the sine square of angle $x$.
2. In the second term, the fraction should have $x$ square in the denominator. So, the factor $1$ in the second term can be written as the quotient of $x$ square divided by the $x$ square.

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

There are two fractions in each term and they involved in the multiplication. So, multiply the fractions in every term to find their products in the expression.

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

The places of the factors in both numerator and denominator of both terms can be changed, as per the commutative property of the multiplication, but keep the denominator of the second term as it is.

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

Now, multiply the expressions in both numerator and denominator of both terms to find their products mathematically.

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

The two fractions are connected by a minus symbol and they both have same denominators. Therefore, the difference between them can be evaluated by the subtraction rule of the fractions having same denominators.

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

Methods

Fundamental Limit method

L’Hôpital’s rule

Due to the indeterminate form, the limit of the one divided by $x$ square minus one divided by sine squared of $x$ can be evaluated by the l’hospital’s rule as the value of $x$ tends to zero.

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