Question:
$\displaystyle \lim_{x\to0}\left(\frac{\tan \,x}{\sqrt{2x +4} - 2}\right)$ is equal to
$\displaystyle \lim_{x\to0}\left(\frac{\tan \,x}{\sqrt{2x +4} - 2}\right)$ is equal to
Updated On: May 30, 2022
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Solution and Explanation
$\displaystyle\lim_{x \rightarrow 0}\left(\frac{\tan \,x}{\sqrt{2 x+4}-2}\right) \,\,\,\left[\frac{0}{0}\right]$ form
Applying L' Hopitals' rule, we get
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{\sec ^{2} x}{\frac{2}{2x\sqrt{2x +4}}-0}\right)$
$=\displaystyle\lim _{x \rightarrow 0}\left((\sqrt{2 x+4}) \sec ^{2} x\right)$
$=\sqrt{2 \times 0+4} \times 1=2$
Applying L' Hopitals' rule, we get
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{\sec ^{2} x}{\frac{2}{2x\sqrt{2x +4}}-0}\right)$
$=\displaystyle\lim _{x \rightarrow 0}\left((\sqrt{2 x+4}) \sec ^{2} x\right)$
$=\sqrt{2 \times 0+4} \times 1=2$
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Concepts Used:
Limits of Trigonometric Functions
Assume a is any number in the general domain of the corresponding trigonometric function, then we can explain the following limits.
We know that the graphs of the functions y = sin x and y = cos x detain distinct values between -1 and 1 as represented in the above figure. Thus, the function is swinging between the values, so it will be impossible for us to obtain the limit of y = sin x and y = cos x as x tends to ±∞. Hence, the limits of all six trigonometric functions when x tends to ±∞ are tabulated below:
