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Question: How do you find the limit of \[\dfrac{\tan 2x}{5x}\] as \[x\] approaches to \[0\]?...

How do you find the limit of tan⁡2x5x\dfrac{\tan 2x}{5x} as xx approaches to 00?

Explanation

Solution

From the question given, we have been asked to find the limit of tan⁡2x5x\dfrac{\tan 2x}{5x} as xx approaches to 00. We can solve the given question by transforming the given function into simplified form and then we have to apply the limits with given condition to the given function.

Complete step by step answer:
From the question given, the given function is tan⁡2x5x\dfrac{\tan 2x}{5x}
As we have already discussed earlier, we have to transform the given equation.
We know that tan⁡x\tan x can be written as sin⁡xcos⁡x\dfrac{\sin x}{\cos x}.
By applying this to the given question, we can write it as,
Therefore, the given function can be written as ⇒tan⁡2x5x=sin⁡(2x)5xcos⁡(2x)\Rightarrow \dfrac{\tan 2x}{5x}=\dfrac{\sin \left( 2x \right)}{5x\cos \left( 2x \right)}
We can also write it as ⇒tan⁡2x5x=sin⁡(2x)2x(25cos⁡(2x))\Rightarrow \dfrac{\tan 2x}{5x}=\dfrac{\sin \left( 2x \right)}{2x}\left( \dfrac{2}{5\cos \left( 2x \right)} \right)
Now, we have to apply the limits to the above simplified function with the given conditions in the question.
By applying the limits to the above simplified function with the given conditions in the question, we get ⇒lim⁡x→0tan⁡2x5x=lim⁡x→0sin⁡(2x)2x×lim⁡x→0(25cos⁡(2x))\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\tan 2x}{5x}=\displaystyle \lim_{x \to 0}\dfrac{\sin \left( 2x \right)}{2x}\times \displaystyle \lim_{x \to 0}\left( \dfrac{2}{5\cos \left( 2x \right)} \right)
Simplify further the above limits to get the simplified answer.
By simplifying further the above limits, we get ⇒lim⁡x→0tan⁡2x5x=1×25=25\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\tan 2x}{5x}=1\times \dfrac{2}{5}=\dfrac{2}{5}
Hence, we got the simplified answer for the given question.

Note: We should be well aware of the concept of limits. Also, we should be very careful while simplifying the given function. We should have to simplify the given function so that it will be easy to apply the limits after simplifying the given question. Also, we should be very careful while doing the calculation part. Also, we should be very careful while applying the limits with given conditions to the given function. Similarly we can find the limit of any trigonometric ratios. There are many trigonometric limit formulae which we can use in this type of questions like lim⁡x→0sin⁡(ax)ax=1\displaystyle \lim_{x \to 0}\dfrac{\sin \left( ax \right)}{ax}=1 .