Math Doubts

Evaluate $\large \displaystyle \lim_{x \to a}$ $\dfrac{\cos \sqrt{x} -\cos \sqrt{a}}{x-a}$

$\sqrt{x}$ and $\sqrt{a}$ are two angles of two right angled triangles. The quotient of subtraction of cos of angle $\sqrt{a}$ from cos angle $\sqrt{x}$ by the subtraction of $a$ from $x$ has to be evaluated when the value of $x$ tends to $a$.

$\large \displaystyle \lim_{x \to a}$ $\dfrac{\cos \sqrt{x} -\cos \sqrt{a}}{x-a}$

Firstly, substitute $x = a$ to test the functionality of this function at $x$ is equal to $a$.

$= \dfrac{\cos \sqrt{a} -\cos \sqrt{a}}{a-a}$

$= \dfrac{0}{0}$

It is an indeterminate form and it is not the required solution for this problem. So, the problem has to be solved in another approach mathematically.

Step: 1

The numerator contains cosine functions and they are connected by a minus sign. They can be transformed in another form by using sum to product form transformation trigonometric identity.

$= \large \displaystyle \lim_{x \to a}$ $\dfrac{-2\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg] \sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{x-a}$

Step: 2

The angle of both sine functions are in root form. So, change the denominator to be appeared in same form.

$= \large \displaystyle \lim_{x \to a}$ $\dfrac{-2\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg] \sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{{(\sqrt{x})}^2-{(\sqrt{a})}^2}$

The subtraction of squares of two terms can be expressed as the product of sum of them and subtraction of them.

$= \large \displaystyle \lim_{x \to a}$ $\dfrac{-2\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg] \sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})(\sqrt{x}-\sqrt{a})}$

Step: 3

The denominator is modified almost same as the angles of the both sine functions. Now, split the whole function into two multiplying factors but keep the multiplying factor in denominator under the sine function whose angle is similar to it.

$= \large \displaystyle \lim_{x \to a}$ $\dfrac{-2\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{(\sqrt{x}-\sqrt{a})}$

The limit condition belongs to both multiplying factors. So, it can be applied to both of them.

$= \large \displaystyle \lim_{x \to a}$ $\dfrac{-2\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{(\sqrt{x}-\sqrt{a})}$

Step: 4

If you submit $x = a$ in first multiplying factor, it is not equal to zero. So, don’t touch the first multiplying factor. If you do it for second multiplying factor, it becomes indeterminate. The second multiplying factor is in the form of lim x tends to 0 sinx/x formula. Hence, adjust the second multiplying factor to be in the same of the formula.

$= -2 \large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{(\sqrt{x}-\sqrt{a}) \times 1 }$

$= -2 \large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{(\sqrt{x}-\sqrt{a}) \times \dfrac{2}{2} }$

$= -2 \large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{ 2 \times \dfrac{\sqrt{x}-\sqrt{a}}{2}}$

$= -2 \large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{1}{2} \times \dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{\dfrac{\sqrt{x}-\sqrt{a}}{2}}$

$= -2 \times \dfrac{1}{2} \times \large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{\dfrac{\sqrt{x}-\sqrt{a}}{2}}$

$= -\dfrac{2}{2} \times \large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{\dfrac{\sqrt{x}-\sqrt{a}}{2}}$

$\require{cancel} = -\dfrac{\cancel{2}}{\cancel{2}} \times \large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{\dfrac{\sqrt{x}-\sqrt{a}}{2}}$

$= -\large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{\dfrac{\sqrt{x}-\sqrt{a}}{2}}$

Step: 5

The second multiplying factor is not completely transformed in the form the limit formula $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{\sin x}{x}$. It can be applied only if the second multiplying factor is successfully transformed in the form of this formula.

If $x \to a \implies \sqrt{x} \to \sqrt{a}$. Then, $\sqrt{x}-\sqrt{a} \to 0$. Finally, $\dfrac{\sqrt{x}-\sqrt{a}}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{\sqrt{x}-\sqrt{a}}{2} \to 0$.

It is proved that if $x \to a$ then $\dfrac{\sqrt{x}-\sqrt{a}}{2} \to 0$. Now change the limit of the second function.

$= -\large \displaystyle \lim_{x \to a}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}+\sqrt{a}}{2}\Bigg]}{(\sqrt{x}+\sqrt{a})}$ $\times$ $\displaystyle \large \lim_{\small \dfrac{\sqrt{x}-\sqrt{a}}{2} \displaystyle \to 0}$ $\dfrac{\sin \Bigg[\dfrac{\sqrt{x}-\sqrt{a}}{2}\Bigg]}{\dfrac{\sqrt{x}-\sqrt{a}}{2}}$

Now, the second multiplying function is completed the transformation in the form our formula.

Step: 6

Substitute $x = a$ in the first multiplying factor and the value of the second multiplying factor is $1$ as per the $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{\sin x}{x}$ rule.

$= -\dfrac{\sin \Bigg[\dfrac{\sqrt{a}+\sqrt{a}}{2}\Bigg]}{(\sqrt{a}+\sqrt{a})}$ $\times$ $1$

$= -\dfrac{\sin \Bigg[\dfrac{2\sqrt{a}}{2}\Bigg]}{2\sqrt{a}}$

$\require{cancel} = -\dfrac{\sin \Bigg[\dfrac{\cancel{2}\sqrt{a}}{\cancel{2}}\Bigg]}{2\sqrt{a}}$

$\therefore \,\,\,\,\,\,$ $\large \displaystyle \lim_{x \to a}$ $\dfrac{\cos \sqrt{x} -\cos \sqrt{a}}{x-a}$ $=$ $- \, \dfrac{\sin \sqrt{a}}{2\sqrt{a}}$

Therefore, it is the required solution for this limits problem in calculus mathematics.



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