$x$ 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 $0$.
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.
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1+1-\cos{x}}{x^2}$
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \Bigg[\dfrac{e^{x^2}-1}{x^2}+\dfrac{1-\cos{x}}{x^2}\Bigg]$
Use sum law of limits to apply limit value to both functions.
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\cos{x}}{x^2}$
The numerator which contains trigonometric function can be simplified by angle to half angle transforming trigonometric identity.
$1-\cos{\theta} = 2\sin^2{\Bigg(\dfrac{\theta}{2}\Bigg)}$
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{2\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{x^2}$
$= \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $2\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{\Bigg(\dfrac{x}{2}\Bigg)}}{x^2}$
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, $\dfrac{x}{2} \to 0$
$= \displaystyle \large \lim_{x^2 \,\to\, 0} \normalsize \dfrac{e^{x^2}-1}{x^2}$ $+$ $\dfrac{1}{2} {\Bigg[\displaystyle \large \lim_{\frac{x}{2} \,\to\, 0} \normalsize \dfrac{\sin{\Bigg(\dfrac{x}{2}\Bigg)}}{\Bigg(\dfrac{x}{2}\Bigg)} \Bigg]}^2$
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.
$= 1+\dfrac{1}{2} \times {(1)}^2$
$= 1+\dfrac{1}{2} \times 1$
$= 1+\dfrac{1}{2}$
$= \dfrac{2+1}{2}$
$= \dfrac{3}{2}$
List of most recently solved mathematics problems.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.