Question

Find the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$
$A)\dfrac{5}{4}$
$B)3$
$C)\dfrac{3}{2}$
$D)2$

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
Since limiting x to zero will be the value of the limiting function and that value will be used to simplify the terms given in this problem.
Also, while applying the limiting functions, we must need to know about the L-hospital rules; which is the zero divides zero or infinite divides infinite terms so the values only exist when we use differentiation after that only we must apply to the constant (we must differentiate one or more than one times to occur this).
Formula used: ${e^{{x^2}}} = 1 + {x^2} + \dfrac{{{x^4}}}{{2!}} + ...$and $\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ...$

Complete step by step answer:
Let from the given question we have $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$and we need to find the value after applying the limiting functions.
First, change the given terms concerning derive as simple like\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}}}{{{{(\dfrac{{\sin x}}{{{x^2}}})}^2}}}\] multiplying and dividing the terms by x power two; Now we are going to apply the values that given above.
Since ${e^{{x^2}}} = 1 + {x^2} + \dfrac{{{x^4}}}{{2!}} + ...$and $\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ...$apply these terms in the above derivation.
Thus, we get \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}}}{{{{(\dfrac{{\sin x}}{{{x^2}}})}^2}}} = \dfrac{{\dfrac{{1 + {x^2} + \dfrac{{{x^4}}}{{2!}} + ... - (1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ...)}}{{{x^2}}}}}{{{{(\dfrac{{\sin x}}{{{x^2}}})}^2}}}\]
While before applying the limit we first simplify the derivations as $1 + {x^2} + \dfrac{{{x^4}}}{{2!}} + ... - (1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ...) = 1 + \dfrac{1}{{2!}}$
Since to apply the limit function we have to convert the given set of terms into solvable terms with the limit so that only we can apply the limit without use of the L-hospital rule and canceling the common terms and also applying the limiting function for the x power two; (the denominator will be zero after applying the limit to x power zero).
Hence, we get$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}} = 1 + \dfrac{1}{{2!}} = \dfrac{3}{2}$
So, the correct answer is “Option C”.

Note: We can also able to solve this problem by using the L-hospital rule (differentiation concerning x or any values and we have to differentiation the only up to the constant values occur like non-zeros after applying the limit values), $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$which is in zero-by-zero form and we have to differentiate concerning x three times after that we get the exact value the option$C)\dfrac{3}{2}$ is correct.