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Evaluate $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$

$x$ is a variable but also represents an angle of the right angled triangle. The exponential function and trigonometric function formed a special algebraic trigonometric function. It is required to find the value of this function as $x$ approaches $0$.

Separate the functions

Exponential function in algebraic form and trigonometric function formed a special function. The combination of them creates a problem in simplification. So, it is good idea to separate both functions. Add one and subtract it by one in numerator before separating them.

$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1+1-\cos{x}}{x^2}$

$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \Bigg[\dfrac{e^{x^2}-1}{x^2}+\dfrac{1-\cos{x}}{x^2}\Bigg]$

Simplify limit of sum of the functions

Use sum law of limits to apply limit value to both functions.

$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\cos{x}}{x^2}$

Simplify the trigonometric function part

The numerator which contains trigonometric function can be simplified by angle to half angle transforming trigonometric identity.

$1-\cos{\theta} = 2\sin^2{\Bigg(\dfrac{\theta}{2}\Bigg)}$

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

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

Transform functions as Limit rules

Now, try to change each function as a rule of limits and then the rules of limits can be applied to find the value of the function as $x$ approaches $0$.

The first limit of the function is almost in the form of $\displaystyle \large \lim_{x \,\to\, 0} \dfrac{e^x-1}{x}$. In this case, the exponential function and denominator contains $x^2$ but limit value is $x$ tends to $0$. So, it should be changed to apply this rule.

If $x \,\to\, 0$, then $x^2 \,\to\, {(0)}^2$. Therefore, $x^2 \,\to\, 0$

The second function is almost same as the $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x}$ rule. So, try to change it mathematically.

$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{4 \times \dfrac{x^2}{4}}$

$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{4 \times {\Bigg(\dfrac{x}{2}\Bigg)}^2}$

$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1}{4} \times \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{{\Bigg(\dfrac{x}{2}\Bigg)}^2}$

$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2 \times \dfrac{1}{4} \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{{\Bigg(\dfrac{x}{2}\Bigg)}^2}$

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

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

$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} \displaystyle \large \lim_{x \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{\Bigg(\dfrac{x}{2}\Bigg)}}{\Bigg(\dfrac{x}{2}\Bigg)} \Bigg]}^2$

$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} {\Bigg[\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{\Bigg(\dfrac{x}{2}\Bigg)}}{\Bigg(\dfrac{x}{2}\Bigg)} \Bigg]}^2$

The denominator of the second function is adjusted same as the angle of the sin function. But, the limit value is different. If $x \,\to\, 0$, then $\dfrac{x}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{x}{2} \to 0$

Obtain the Solution of Limit Problem

$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} {\Bigg[\displaystyle \large \lim_{\frac{x}{2} \,\to\, 0} \normalsize \dfrac{\sin{\Bigg(\dfrac{x}{2}\Bigg)}}{\Bigg(\dfrac{x}{2}\Bigg)} \Bigg]}^2$

According to lim x tends 0 (e^x-1)/x formula, the value of the first function is $1$. Similarly, the value of the second function is also $1$ as per limit x approaches 0 sinx/x rule.

$= 1+\dfrac{1}{2} \times {(1)}^2$

$= 1+\dfrac{1}{2} \times 1$

$= 1+\dfrac{1}{2}$

$= \dfrac{2+1}{2}$

$= \dfrac{3}{2}$

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