How do you find the limit of \[\dfrac{\tan 2x}{5x}\] as \[x\] approaches to \[0\]?
Answer
597.9k+ views
Hint: From the question given, we have been asked to find the limit of \[\dfrac{\tan 2x}{5x}\] as \[x\] approaches to \[0\]. 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 \[\dfrac{\tan 2x}{5x}\]
As we have already discussed earlier, we have to transform the given equation.
We know that \[\tan x\] can be written as \[\dfrac{\sin x}{\cos x}\].
By applying this to the given question, we can write it as,
Therefore, the given function can be written as \[\Rightarrow \dfrac{\tan 2x}{5x}=\dfrac{\sin \left( 2x \right)}{5x\cos \left( 2x \right)}\]
We can also write it as \[\Rightarrow \dfrac{\tan 2x}{5x}=\dfrac{\sin \left( 2x \right)}{2x}\left( \dfrac{2}{5\cos \left( 2x \right)} \right)\]
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 \[\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\tan 2x}{5x}=\displaystyle \lim_{x \to 0}\dfrac{\sin \left( 2x \right)}{2x}\times \displaystyle \lim_{x \to 0}\left( \dfrac{2}{5\cos \left( 2x \right)} \right)\]
Simplify further the above limits to get the simplified answer.
By simplifying further the above limits, we get \[\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\tan 2x}{5x}=1\times \dfrac{2}{5}=\dfrac{2}{5}\]
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 \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( ax \right)}{ax}=1\] .
Complete step by step answer:
From the question given, the given function is \[\dfrac{\tan 2x}{5x}\]
As we have already discussed earlier, we have to transform the given equation.
We know that \[\tan x\] can be written as \[\dfrac{\sin x}{\cos x}\].
By applying this to the given question, we can write it as,
Therefore, the given function can be written as \[\Rightarrow \dfrac{\tan 2x}{5x}=\dfrac{\sin \left( 2x \right)}{5x\cos \left( 2x \right)}\]
We can also write it as \[\Rightarrow \dfrac{\tan 2x}{5x}=\dfrac{\sin \left( 2x \right)}{2x}\left( \dfrac{2}{5\cos \left( 2x \right)} \right)\]
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 \[\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\tan 2x}{5x}=\displaystyle \lim_{x \to 0}\dfrac{\sin \left( 2x \right)}{2x}\times \displaystyle \lim_{x \to 0}\left( \dfrac{2}{5\cos \left( 2x \right)} \right)\]
Simplify further the above limits to get the simplified answer.
By simplifying further the above limits, we get \[\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\tan 2x}{5x}=1\times \dfrac{2}{5}=\dfrac{2}{5}\]
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 \[\displaystyle \lim_{x \to 0}\dfrac{\sin \left( ax \right)}{ax}=1\] .
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is cell theory Who formulated it class 11 biology CBSE

Phyllotaxy is the arrangement of ALeaflets BLeaves class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

The symbiotic association of fungi and algae is called class 11 biology CBSE

Cell theory was formulated by A Schleiden and Schwann class 11 biology CBSE

