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Question
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.
Options
0.00
1.00
2.00
3.00
MCQ
Fill in the Blanks
Solution
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is 1.00.
Explanation:
`lim_(x→0) (sinx)^2/((e^(tan^2x) - 1))`
= `lim_(x→0)(sinx/x)^2 xx (tan^2x)/((e^(tan^2x) - 1)) xx (x/tanx)^2` = 1
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Concept of Limits
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